Electrostatics

Electric Charge 

Like mass, electric charge is an intrinsic property of protons and electrons, and only two types of charge have been discovered positive and negative. A proton has a positive charge, and an electron has a negative charge. A neutron has no net electric charge. The magnitude of the charge on the proton exactly equals the magnitude of the charge on the electron. The proton carries a charge \(+e\) and the electron carries a charge \(-e\). The SI unit of charge is coulomb (C) and \(e\) has the value
\(
e=1.6 \times 10^{-19} \mathrm{C}
\)

Electric charge (symbol \(q\), sometimes \(Q\) ) is a physical property of matter (like protons and electrons) that causes it to experience a force when placed in an electromagnetic field. Electric field of a positive and a negative point charge is shown in above figure. Electric charge is a conserved property: the net charge of an isolated system, the quantity of positive charge minus the amount of negative charge, cannot change. 

Quantisation of Charge

When we say the electric charge is quantized, it means that charge exists in discrete, indivisible units (like “packets”) rather than as a continuous, infinitely divisible quantity, specifically as integer multiples of the elementary charge (the charge of an electron).
Charge on any body is always an integral multiple of elementary charge (\(\pm e\))
\(
{Q} =q = \pm {n e} \quad \text { where, } \quad n=1,2,3 \ldots
\)
where \(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\), Charge on any body can never be \(\left(\frac{1}{3} e\right), 1.5 e\), etc.

Charge is conserved

The total charge of an isolated system (An isolated system, in physics, is a system that doesn’t exchange energy or matter with its surroundings, meaning its total energy and mass remain constant) is always conserved. It is not possible to create or destroy net charge carried by any isolated system. It can only be transferred from one body to another body. Pair production and pair annihilation are two examples of conservation of charge.

Charging of a Body

Mainly there are the following three methods of charging a body :

Charging by Friction (rubbing):  This involves rubbing two different materials together, causing a transfer of electrons from one material to the other, resulting in one material becoming positively charged and the other becoming negatively charged.
Example: When a glass rod is rubbed with a silk cloth, the glass rod acquires some positive charge and the silk cloth acquires negative charge by the same amount.

Charging by Conduction (or Contact): This method involves bringing a charged object into contact with an uncharged object. The charge from the charged object will then be transferred to the uncharged object, making it charged.
Example: Touching a metal doorknob after walking on a carpet, which causes electrons to flow from your charged body to the doorknob, resulting in both your hand and the doorknob having a negative charge.

Charging by Induction:
Charging by induction is a method of charging an object without direct contact, achieved by bringing a charged object near a neutral object, which then polarizes and allows for charge separation, resulting in the neutral object becoming charged.
Example: Bringing a negatively charged rod near a metal sphere will cause the near end of the sphere to become positively charged and the far end to become negatively charged. If the sphere is grounded and the rod is removed, the sphere will retain a negative charge.

Example 1: If we comb our hair on a dry day and bring the comb near small pieces of paper, the comb attracts the pieces, why?

Solution: This is an example of frictional electricity and induction. When we comb our hair, it gets positively charged by rubbing. When the comb is brought near the pieces of paper some of the electrons accumulate at the edge of the paper piece which is closer to the comb. At the farther end of the piece, there is deficiency of electrons and hence, positive charge appears there. Such a redistribution of charge in a material, due to presence of a nearby charged body is called inducion. The comb exerts larger attraction on the negative charges of the paper piece as compared to the repulsion on the positive charge. This is because the negative charges are closer to the comb. Hence, there is a net attraction between the comb and the paper piece.

Example 2: Does the attraction between the comb and the piece of papers last for longer period of time?

Solution: No, because the comb loses its net charge after some time. The excess charge of the comb transfers to earth through our body after some time.

Example 3: Does in charging the mass of a body change?

Solution: Yes, as charging a body means addition or removal of electrons and electron has a mass.

Important points regarding electric charge:

• Like charges repel each other and unlike charges attract each other.
• Charge is a scalar quantity as it has magnitude but no direction. It can be of two types as positive and negative. When some electrons are removed from the atom, it acquires a positive charge and when some electrons are added to the atom, it acquires a negative charge.
• Charge can be transferred from one body to another. Charge transfer, like rubbing a balloon on your hair, occurs when electrons move from one object to another, leaving one object with a net positive charge and the other with a net negative charge.
• During any process, the net electric charge of an isolated system remains constant or we can say that charge is conserved. Pair production and pair annihilation are two examples of conservation of charge.
• A stationary (at rest) charged particle creates an electric field around it, which is a static field.
• When a charged particle moves at a constant velocity, it generates both electric and magnetic fields. However, this motion does not result in the emission of electromagnetic radiation or energy. 
• Acceleration, or a change in velocity, is what triggers the emission of electromagnetic radiation. The accelerated charge not only produces electric and magnetic fields but also radiates energy in the form of electromagnetic waves, which are disturbances in the electric and magnetic fields that propagate through space. For example, When electrons accelerate (e.g., in a radio antenna or a cathode ray tube), they emit electromagnetic radiation, which can be detected as radio waves or light, respectively.

Example 4: How many electrons are there in one coulomb of negative charge?

Solution: The negative charge is due to the presence of excess electrons, since they carry negative charge. Because an electron has a charge whose magnitude is \(e=1.6 \times 10^{-19} \mathrm{C}\), the number of electrons is equal to the charge \(q\) divided by the charge \(e\) on each electron. Therefore, the number \(n\) of electrons is
\(
n=\frac{q}{e}=\frac{1.0}{1.6 \times 10^{-19}}=6.25 \times 10^{18}
\)

Coulomb’s Law (Vacuum)

The electrostatic force of interaction between two static point electric charges is directly proportional to the product of the charges, inversely proportional to the square of the distance between them and acts along the straight line joining the two charges.

The magnitude of the electric force exerted by a charge \(q_1\) on another charge \(q_2\) a distance \(r\) away is thus, given by (according to Coulomb’slaw),
\(
{F} \propto {q}_1 {q}_2 \text { and } {F} \propto \frac{1}{{r}^2}
\)
\(
F=\frac{k\left|q_1 q_2\right|}{r^2}
\)
The constant \(k\) is usually put as \(k=\frac{1}{4 \pi \varepsilon_0}\), where \(\varepsilon_0\) is called the permittivity of free space (vacuum) and has the value \(\varepsilon_0=8.854 \times 10^{-12} \mathrm{C}^2 / \mathrm{N}-\mathrm{m}^2\). For all practical purposes, we will take \(\frac{1}{4 \pi \varepsilon_0} \simeq 9 \times 10^9 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}^2\).

Coulomb’s Law (In a medium)

The presence of a medium between the charges reduces the electrostatic force compared to the force in a vacuum. This reduction is quantified by the permittivity of the medium ( \(\varepsilon\) ), which is a material property. In SI units, Coulomb’s constant (\(k\)) is often expressed as \(1 /(4 \pi \varepsilon)\), where \(\varepsilon\) is the permittivity of the medium.

The relative permittivity (\(\varepsilon_r\)), also known as the dielectric constant, is the ratio of the permittivity of the medium to the permittivity of a vacuum ( \(\varepsilon_r=\varepsilon / \varepsilon_0\) ). The dielectric constant is a dimensionless quantity that indicates how much the force between charges is reduced in a particular medium compared to a vacuum. A higher dielectric constant means a greater reduction in the electrostatic force.
\(F=\frac{1}{4 \pi \varepsilon} \frac{\left|q_1 q_2\right|}{r^2}\)

Coulomb’s law in vector form

Consider two point charges \(q_1\) and \(q_2\) placed in vacuum at a distance \(r\) from each other, attract each other.

\(
\begin{aligned}
&\text { In vector form, Coulomb’s law may be expressed as }\\
&\mathbf{F}_{12}=\text { Force on charge } q_1 \text { due to } q_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2} \hat{\mathbf{r}}_{12}
\end{aligned}
\)
\(\text { where, } \hat{\mathbf{r}}_{12}=\frac{\mathbf{r}_{12}}{\left|\mathbf{r}_{12}\right|}=\frac{\mathbf{r}_{12}}{r} \text { is a unit vector in the direction from } q_1 \text { to } q_2 \text {. }\)
Similarly, \(\mathbf{F}_{21}=\) force on charge \(q_2\) due to \(q_1\)
\(
=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2} \hat{\mathbf{r}}_{21}
\)
where, \(\hat{\mathbf{r}}_{21}=\frac{\mathbf{r}_{21}}{\left|\mathbf{r}_{12}\right|}\) is a unit vector in the direction from \(q_2\) to \(q_1\).

Important Points

• If \(q_1\) and \(q_2\) are of same sign ; \(q_1 q_2>0\) then \(\mathbf{F}_{12}\) is along \(\mathbf{r}_{12}\), which denotes repulsion
• If \(q_1\) and \(q_2\) are of opposite sign; \(q_1 q_2<0\) then \(\mathbf{F}_{12}\) is along \(-\mathbf{r}_{12}\) or \(\mathbf{r}_{21}\), which denotes attraction
• The electric force is an action-reaction pair, i.e. the two charges exert equal and opposite forces on each other. Thus, Coulomb’s law obeys Newton’s third law.
  \(
  \mathbf{F}_{12}=-\mathbf{F}_{21}
  \)
• The electric force is conservative in nature, meaning the work done by it depends only on the initial and final positions, not the path taken.
• Electric forces are significantly stronger than gravitational forces, especially at the atomic level, because the electrostatic constant \((k)\) is much larger than the gravitational constant (\(G\)), and because electric charges can be positive or negative, leading to both attraction and repulsion, while gravity only involves attraction.
  The electric force is much stronger than gravitational force, \(F_e \gg F_g\). As value of \(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}^2\), which is much more than the value of gravitational constant,
  \(
  G=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 \mathrm{~kg}^{-2} .
  \)

Coulomb’s law in terms of position vector

Let the position vector of charges \(q_1\) and \(q_2\) be \(\vec{r}_1\) and \(\vec{r}_2\) respectively.
We denote the force \(q_1\) due to \(q_2\) by \(\vec{F_{12}}\) and the force on \(q_2\) due to \(q_1\) by \(\vec{F}_{21}\).
Here \(\quad \vec{r}_1+\vec{r}_{21}=\vec{r_2}\)
\(\quad \vec{r_{21}}=\vec{r}_2-\vec{r}_1\)
also \(\vec{r_2}+\vec{r_{12}}=\vec{r_1}\)
or \(\quad \vec{r_{12}}=\vec{r}_1-\vec{r}_2\)
and \(\vec{r_{12}}=-\vec{r}_{21}\)
The magnitude of vectors \(\vec{r}_{12}\) and \(\vec{r}_{21}\) is denoted by \(\left|\vec{r}_{12}\right|\) and \(\left|\vec{r}_{21}\right|\) and their unit vectors are given as 
\(
\hat{r}_{12}=\frac{\vec{r}_{12}}{\left|\vec{r}_{12}\right|} \quad \text { and } \quad \hat{r}_{21}=\frac{\vec{r}_{21}}{\left|\vec{r}_{21}\right|}
\)
So the force \(\vec{F}_{12}\) and \(\vec{F}_{21}\) can be given as
\(
\vec{F}_{12}=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{\left|\vec{r}_{12}\right|^2} \hat{r}_{12}
\)
\(
\vec{F}_{21}=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{\left|\vec{r}_{21}\right|^2} \hat{r}_{21}
\)

Example 5: Two charges \({q}_1\) and \({q}_2\) of magnitude 3 C and 4 C are kept on two corners of a right-angled triangle with sides \(3 \mathrm{~m}, 4 \mathrm{~m}\) and \(5 \mathrm{~m}\). Find the force acting on them.

Solution: We have \(q_1=3 \mathrm{C}\) and \({q}_2=4 \mathrm{C}\), and \(r\) equals to 5 m which is the shortest distance between both the charges. Now substituting values accordingly in the relation
\(
\begin{aligned}
& {F}={k} q_1 {q}_2 / {r}^2 \\
& {~F}=9 \times 10^9 \times 3 \times 4 / 5^2 \\
& {~F}=4.32 \times 10^9 \text { Newton }
\end{aligned}
\)

Force between multiple charges (Superposition principle)

According to the principle of superposition, “total force on a given charge due to the number of charges is the vector sum of individual forces acting on that charge due to all the charges”. The individual forces are unaffected due to the presence of other charges.

Suppose a system contains \(n\) point charges \(q_1, q_2, \ldots, q_n\). Then, by the principle of superposition, the force on \(q_1\) due to all the other charges is given by
\(
\begin{aligned}
\mathbf{F}_1 & =\mathbf{F}_{12}+\mathbf{F}_{13}+\ldots \ldots+\mathbf{F}_{1 n} \\
\mathbf{F}_1 & =\frac{q_1}{4 \pi \varepsilon_0}\left[\frac{q_2 \hat{\mathbf{r}}_{12}}{r_{12}^2}+\frac{q_3 \hat{\mathbf{r}}_{13}}{r_{13}^2}+\ldots \ldots \cdot \frac{q_n \hat{\mathbf{r}}_{1 n}}{r_{1 n}^2}\right] \\
& =\frac{q_1}{4 \pi \varepsilon_0} \sum_{i=2}^n \frac{q_i \hat{\mathbf{r}}_{1 i}}{r_{1 i}^2}
\end{aligned}
\)

Important Points:

• Suppose the position vectors of two charges \(q_1\) and \(q_2\) are \(\mathbf{r}_1\) and \(\mathbf{r}_2\), then electric force on charge \(q_1\) due to charge \(q_2\) is,
  \(
  \mathrm{F}_{12}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{\left|\mathbf{r}_1-\mathbf{r}_2\right|^3}\left(\mathbf{r}_1-\mathbf{r}_2\right)
  \)
  Similarly, electric force on \(q_2\) due to charge \(q_1\) is
  \(
  \mathbf{F}_{21}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{\left|\mathbf{r}_2-\mathbf{r}_1\right|^3}\left(\mathbf{r}_2-\mathbf{r}_1\right)
  \)
  Here, \(q_1\) and \(q_2\) are to be substituted with sign. \(\mathbf{r}_1=x_1 \hat{\mathbf{i}}+y_1 \hat{\mathbf{j}}+z_1 \hat{\mathbf{k}}\) and \(\mathbf{r}_2=x_2 \hat{\mathbf{i}}+y_2 \hat{\mathbf{j}}+z_2 \hat{\mathbf{k}}\) where \(\left(x_1, y_1, z_1\right)\) and \(\left(x_2, y_2, z_2\right)\) are the coordinates of charges \(q_1\) and \(q_2\).

Lami’s theorem

According to this theorem, “if three concurrent forces \(F_1, F_2\) and \(F_3\) as shown in Figure above are in equilibrium or if \(F_1+F_2+F_3=0\), then
\(
\frac{F_1}{\sin \alpha}=\frac{F_2}{\sin \beta}=\frac{F_3}{\sin \gamma}
\)

Example 6: Two identical balls each having a density \(\rho\) are suspended from a common point by two insulating strings of equal length. Both the balls have equal mass and charge. In equilibrium, each string makes an angle \(\theta\) with vertical. Now, both the balls are immersed in a liquid. As a result, the angle \(\theta\) does not change. The density of the liquid is \(\sigma\). Find the dielectric constant of the liquid.

Solution: Each ball is in equilibrium under the following three forces
(i) tension,
(ii) electric force and
(iii) weight.

So, by applying Lami’s theorem,

In the liquid, \(F_e^{\prime}=\frac{F_e}{K}\) where, \(K=\) dielectric constant of liquid and \(\quad w^{\prime}=w\) – upthrust
Applying Lami’s theorem in vacuum,
\(
\frac{w}{\sin \left(90^{\circ}+\theta\right)}=\frac{F_e}{\sin \left(180^{\circ}-\theta\right)} \text { or } \frac{w}{\cos \theta}=\frac{F_e}{\sin \theta} \dots(i)
\)
\(
\text { Similarly in liquid, } \frac{w^{\prime}}{\cos \theta}=\frac{F_e^{\prime}}{\sin \theta} \dots(ii)
\)
On dividing Eq. (i) by Eq. (ii), we get
\(
\frac{w}{w^{\prime}}=\frac{F_e}{F_e^{\prime}} \text { or } K=\frac{w}{w-\text { upthrust }} \quad \left(\text { as } \frac{F_e}{F_e^{\prime}}=K\right)
\)
\(
=\frac{V \rho g}{V \rho g-V \sigma g} \quad(\because V=\text { volume of ball })
\)
or \(\quad K=\frac{\rho}{\rho-\sigma}\)

Example 7: Equal charges each of \(20 \mu \mathrm{C}\) are placed at \(x=0,2,4,8,16 \mathrm{~cm}\) on \(X\)-axis. Find the force experienced by the charge at \(x=2 \mathrm{~cm}\).

Solution: Force on charge at \(x=2 \mathrm{~cm}\) due to charge at \(x=0 \mathrm{~cm}\) and \(x=4 \mathrm{~cm}\) are equal and opposite, so they cancel each other.
The net force on the charge at \(x=2 \mathrm{~cm}\) is the resultant of repulsive forces due to two charges at \(x=8 \mathrm{~cm}\) and \(x=16 \mathrm{~cm}\).
\(
\begin{aligned}
\therefore \quad F & =\frac{q_1 \times q_2}{4 \pi \varepsilon_0}\left[\frac{1}{(0.08-0.02)^2}+\frac{1}{(0.16-0.02)^2}\right] \\
& =9 \times 10^9\left[20 \times 10^{-6}\right]^2\left[\frac{1}{(0.06)^2}+\frac{1}{(0.14)^2}\right]=1.2 \times 10^3 \mathrm{~N}
\end{aligned}
\)

Example 8: Three charges \(q_1=1 \mu C, q_2=-2 \mu C\) and \(q_3=3 \mu C\) are placed on the vertices of an equilateral triangle of side 1.0 m. Find the net electric force acting on charge \(q_1\).

Solution: Charge \(q_2\) will attract charge \(q_1\) (along the line joining them) and charge \(q_3\) will repel charge \(q_1\). Therefore, two forces will act on \(q_1\), one due to \(q_2\) and another due to \(q_3\). Since the force is a vector quantity both of these forces (say \(\mathbf{F}_1\) and \(\mathbf{F}_2\) ) will be added by vector method. Following are two methods of their addition.

Magnitude of force between \(q_1\) and \(q_2\),
\(
\left|\mathbf{F}_1\right|=F_1=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2}
\)
where, \(q_1=1 \mu \mathrm{C}=1 \times 10^{-6} \mathrm{C}\)
and \(\quad q_2=2 \mu \mathrm{C}=2 \times 10^{-6} \mathrm{C}\).
\(
\Rightarrow \quad F_1=\frac{\left(9.0 \times 10^9\right)\left(1.0 \times 10^{-6}\right)\left(2.0 \times 10^{-6}\right)}{(1.0)^2}=1.8 \times 10^{-2} \mathrm{~N}
\)
Similarly, magnitude of force between \(q_1\) and \(q_3\),
\(
\left|\mathbf{F}_2\right|=F_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_3}{r^2}
\)
where, \(q_3=3 \mu \mathrm{C}=3 \times 10^{-6} \mathrm{C}\).
\(
\begin{aligned}
\Rightarrow \quad F_2 & =\frac{\left(9.0 \times 10^9\right)\left(1.0 \times 10^{-6}\right)\left(3.0 \times 10^{-6}\right)}{(1.0)^2} \\
& =2.7 \times 10^{-2} \mathrm{~N}
\end{aligned}
\)
Now, net force, \(\left|\mathbf{F}_{\text {net }}\right|=\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos 120^{\circ}}\)
\(
=\left(\sqrt{(1.8)^2+(2.7)^2+2(1.8)(2.7)\left(-\frac{1}{2}\right)}\right) \times 10^{-2} \mathrm{~N}
\)
\(
\begin{aligned}
& =2.38 \times 10^{-2} \mathrm{~N} \\
\text { and } \tan \alpha & =\frac{F_2 \sin 120^{\circ}}{F_1+F_2 \cos 120^{\circ}} \\
& =\frac{\left(2.7 \times 10^{-2}\right)(0.87)}{\left(1.8 \times 10^{-2}\right)+\left(2.7 \times 10^{-2}\right)\left(-\frac{1}{2}\right)} \\
\text { or } \quad \alpha & =79.2^{\circ}
\end{aligned}
\)
Thus, the net electric force on charge \(q_1\) is \(2.38 \times 10^{-2} \mathrm{~N}\) at an angle \(\alpha=79.2^{\circ}\) with a line joining \(q_1\) and \(q_2\) as shown in the figure.

Alternate:

In this method, let us assume coordinate axes with \(q_1\) at origin as shown in the figure. The coordinates of \(q_1, q_2\) and \(q_3\) in this coordinate system are \((0,0,0),(1 \mathrm{~m}, 0,0)\) and ( \(0.5 \mathrm{~m}, 0.87 \mathrm{~m}, 0\) ), respectively. Now,
\(
\begin{aligned}
& \mathbf{F}_1=\text { Force on } q_1 \text { due to charge } q_2 \\
&=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{\left|\mathbf{r}_1-\mathbf{r}_2\right|^3}\left(\mathbf{r}_1-\mathbf{r}_2\right) \\
&=\frac{\left(9.0 \times 10^9\right)\left(1.0 \times 10^{-6}\right)\left(-2.0 \times 10^{-6}\right)}{(1.0)^3} \\
& \times[(0-1) \hat{\mathbf{i}}+(0-0) \hat{\mathbf{j}}+(0-0) \hat{\mathbf{k}}] \\
&=\left(1.8 \times 10^{-2} \hat{\mathbf{i}}\right) \mathrm{N}
\end{aligned}
\)
\(
\begin{aligned}
&\text { and } \mathbf{F}_2=\text { Force on } q_1 \text { due to charge } q_3\\
&=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_3}{\left|\mathbf{r}_1-\mathbf{r}_3\right|^3}\left(\mathbf{r}_1-\mathbf{r}_3\right)
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{\left(9.0 \times 10^9\right)\left(1.0 \times 10^{-6}\right)\left(3.0 \times 10^{-6}\right)}{(1.0)^3} \\
& \quad \times[(0-0.5) \hat{\mathbf{i}}+(0-0.87) \hat{\mathbf{j}}+(0-0) \hat{\mathbf{k}}] \\
& =(-1.35 \hat{\mathbf{i}}-2.349 \hat{\mathbf{j}}) \times 10^{-2} \mathrm{~N}
\end{aligned}
\)
Therefore, the net force on \(q_1\) is
\(
\begin{aligned}
\mathbf{F} & =\mathbf{F}_1+\mathbf{F}_2 \\
& =(0.45 \hat{\mathbf{i}}-2.349 \hat{\mathbf{j}}) \times 10^{-2} \mathrm{~N} \\
|\mathbf{F}| & =\sqrt{(0.45)^2+(2.349)^2} \times 10^{-2} \mathrm{~N}=2.39 \times 10^{-2} \mathrm{~N}
\end{aligned}
\)
If the net force makes an angle \(\alpha\) from the direction of \(X\)-axis, then
\(
\alpha=\tan ^{-1}\left(\frac{-2.349}{0.45}\right)=-79.2^{\circ}
\)
The negative sign of \(\alpha\) indicates that the net force is directed below the \(X\)-axis.

Electric Field

A charged particle cannot directly interact with another particle kept at a distance. A charge produces something called an electric field in the space around it and this electric field exerts a force on any other charge (except the source charge itself) placed in it. Thus, the region surrounding a charge or distribution of charge in which its electrical effects can be observed is called the electric field of the charge or distribution of charge. Electric field at a point can be defined in terms of either a vector function \(\mathbf{E}=\) called ‘electric field strength‘ or a scalar function \(V\) called ‘electric potential’. The electric field can also be visualised graphically in terms of ‘lines of force’.

Electric field strength

The electric field strength (often called electric field) a point is defined as the electrostatic force \(\mathbf{F}\) per unit positive test charge. Thus, if the electrostatic force experienced by a small test charge \(q_0\) is \(\mathbf{F}\), then field strength at that point is defined as
\(
\mathbf{E}=\lim _{q_0 \rightarrow 0} \frac{\mathbf{F}}{q_0}
\)
The electric field is a vector quantity and its direction is the same as the direction of the electrostatic force \(\mathbf{F}\) on a positive test charge.
The SI unit of the electric field is N/C. The dimensions for \(\mathbf{E}\) is \(\left[\mathrm{MLT}^{-3} \mathrm{~A}^{-1}\right]\). For a positive charge, the electric field will be directed radially outwards from the charge. On the other hand, if the source charge is negative, the electric field vector at each point is directed radially inwards.

Example 9: An electric field of \(10^5 \mathrm{~N} / \mathrm{C}\) points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of \(+2 \mu C\) and \(-5 \mu C\) at this spot?

Solution:

\(
\begin{aligned}
\text { Force on }+2 \mu \mathrm{C} & =q E=\left(2 \times 10^{-6}\right)\left(10^5\right) \\
& =0.2 \mathrm{~N} \quad \text { (due west) }
\end{aligned}
\)
\(
\begin{aligned}
\text { Force on }-5 \mu \mathrm{C} & =\left(5 \times 10^{-6}\right)\left(10^5\right) \\
& =0.5 \mathrm{~N} \quad \text { (due east) }
\end{aligned}
\)

Electric Field due to a Point Charge

The electric field produced by a point charge \(q\) can be obtained in general terms from Coulomb’s law. The magnitude of the force exerted by the charge \(q\) on a test charge \(q_0\) is

\(
F_e=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q q_0}{r^2}
\)
Therefore, the intensity of the electric field at this point is given by
\(
E=\frac{F_e}{q_0}
\)
\(
E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}
\)

If \(q\) is positive, \(\mathbf{E}\) is directed away from \(q\). On the other hand, if \(q\) is negative, then \(\mathbf{E}\) is directed towards \(q\).
Points to remember

• The electric field at a point is a vector quantity. Suppose \(\mathbf{E}_1\) is the field at a point due to a charge \(q_1\) and \(\mathbf{E}_2\) in the field at the same point due to a charge \(q_2\). The resultant field when both the charges are present is
  \(
  \mathbf{E}=\mathbf{E}_1+\mathbf{E}_2
  \)
  If the given charge distribution is continuous, we can use the technique of integration to find the resultant electric field at a point.
• Suppose a charge \(q\) is placed at a point whose position vector is \(\mathbf{r}_q\) and to obtain the electric field at a point \(P\) whose position vector is \(\mathbf{r}_p\), then in vector form the electric field is given by
  \(
  \boldsymbol{E}=\frac{1}{4 \pi \varepsilon_0} \frac{q}{\left|\boldsymbol{r}_p-\boldsymbol{r}_q\right|^3}\left(\boldsymbol{r}_p-\boldsymbol{r}_q\right)
  \)
  Here, \(\boldsymbol{r}_p=x_p \hat{\boldsymbol{i}}+y_p \hat{\boldsymbol{j}}+z_p \hat{\boldsymbol{k}}\)
  and \(\boldsymbol{r}_q=x_q \hat{\boldsymbol{i}}+y_q \hat{\boldsymbol{j}}+z_q \hat{\boldsymbol{k}}\)

Example 10: Two positive point charges \(q_1=16 \mu C\) and \(q_2=4 \mu C\), are separated in vacuum by a distance of 3.0 m. Find the point on the line between the charges where the net electric field is zero.

Solution: Between the charges the two field contributions have opposite directions, and the net electric field is zero at a point (say \(P\) ) where the magnitudes of \(\mathbf{E}_1\) and \(\mathbf{E}_2\) are equal. However, since \(q_2<q_1\), point \(P\) must be closer to \(q_2\), in order that the field of the smaller charge can balance the field of the larger charge.

\(
\text { At } P, E_1=E_2
\)
\(
\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{r_1^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2^2}
\)
\(
\frac{r_1}{r_2}=\sqrt{\frac{q_1}{q_2}}=\sqrt{\frac{16}{4}}=2 \dots(i)
\)
Also,
\(
r_1+r_2=3.0 \mathrm{~m} \dots(ii)
\)
Solving these equations, we get
\(
r_1=2 \mathrm{~m} \text { and } r_2=1 \mathrm{~m}
\)
Thus, the point \(P\) is at a distance of 2 m from \(q_1\) and 1 m from \(q_2\).

Example 11: An infinite number of charges each equal to \(q\) are placed along \(X\)-axis at \(x=1, x=2, x=4, x=8\) and so on. Find the electric field at the point \(x=0\) due to this set up of charges.

Solution: At the point \(x=0\), the electric field due to all the charges are in the same negative \(x\)-direction and hence get added up.
\(
\begin{aligned}
E & =\frac{1}{4 \pi \varepsilon_0}\left[\frac{q}{1^2}+\frac{q}{2^2}+\frac{q}{4^2}+\frac{q}{8^2}+\ldots\right] \\
& =\frac{q}{4 \pi \varepsilon_0}\left[1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots\right] \\
& =\frac{q}{4 \pi \varepsilon_0}\left(\frac{1}{1-1 / 4}\right)=\frac{q}{3 \pi \varepsilon_0}
\end{aligned}
\)
This electric field is along the negative \(X\)-axis.

Electric Field Lines

As we have seen, electric charges create an electric field in the space surrounding them. It is useful to have a kind of “map” that gives the direction and indicates the strength of the field at various places. Field lines, a concept introduced by Michael Faraday, provide us with an easy way to visualize the electric field.

“An electric field line is an imaginary line or curve drawn through a region of space so that its tangent at any point is in the direction of the electric field vector at that point. The relative closeness of the lines at some place gives an idea about the intensity of the electric field at that point.”

The electric field lines have the following properties :

• The tangent to a line at any point gives the direction of \(\mathbf{E}\) at that point. This is also the path on which a positive test charge will tend to move if free to do so.
• Electric field lines always begin on a positive charge and end on a negative charge and do not start or stop in mid-space.
• The number of lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.
• Two lines can never intersect. If it happens then two tangents can be drawn at their point of intersection, i.e. intensity at that point will have two directions which is absurd.
• In a uniform field, the field lines are straight parallel, and uniformly spaced.

• The electric field lines can never form closed loops as a line can never start and end on the same charge.
• Electric field lines also give us an indication of the equipotential surface (the surface which has the same potential)
• Electric field lines always flow from higher potential to lower potential.
• In a region where there is no electric field, lines are absent. This is why inside a conductor (where the electric field is zero) there, cannot be any electric field line.
• Electric lines of force ends or starts normally from the surface of a conductor.

Continuous charge distribution

In most of the cases, we deal with charges having a magnitude greater than the charge of an electron. For this, we can imagine that the charge is spread in a region in a continuous manner. Such a charge distribution is known as continuous charge distribution.
Consider a point charge \(q_0\) lying near a region of continuous charge distribution which is made up of large number of small charges \(d q\) as shown in the figure.

According to Coulomb’s law, the force on a point charge \(q_0\) due to small charge \(d q\) is
\(
\mathbf{F}=\frac{q_0}{4 \pi \varepsilon_0} \int \frac{d q}{r^2} \hat{\mathbf{r}} \dots(i)
\)
where, \(\quad \hat{\mathbf{r}}=\frac{\mathbf{r}}{r}\)
There are three types of continuous charge distribution

Line charge distribution ( \(\lambda\) )

It is a charge distribution along a one-dimensional curve or line, \(L\) in space as shown in the figure.

The charge contained per unit length of the line at any point is called linear charge density. It is denoted by \(\lambda\). i.e.
\(
\lambda=\frac{d q}{d L}
\)
Its SI unit is \(\mathrm{Cm}^{-1}\).
Electric field due to the line charge distribution at the location of charge \(q\) is
\(
\mathbf{E}_L=\frac{1}{4 \pi \varepsilon_0} \int_L \frac{\lambda}{r^2} d L \hat{\mathbf{r}}
\)

Surface charge distribution ( \(\sigma\) )

It is a charge distribution spread over a two-dimensional surface \(S\) in space as shown in the figure.

The charge contained per unit area at any point is called surface charge density. It is denoted by \(\sigma\).
i.e. \(\sigma=\frac{d q}{d S}\)
Its SI unit is \(\mathrm{Cm}^{-2}\).
Electric field due to the surface charge distribution at the location of charge \(q\) is
\(
\mathbf{E}_S=\frac{1}{4 \pi \varepsilon_0} \int_S \frac{\sigma}{r^2} d S \hat{\mathbf{r}}
\)

Volume charge distribution ( \(\rho\) )

It is a charge distribution spread over a three-dimensional volume or region \(V\) of space as shown in the figure.

The charge contained per unit volume at any point is called volume charge density. It is denoted by \(\rho\).
i.e. \(\rho=\frac{d q}{d V}\)
Its SI unit is Coulomb per cubic metre \(\left(\mathrm{Cm}^{-3}\right)\).
Electric field due to the volume charge distribution at the location of charge \(q\) is
\(
\mathbf{E}_V=\frac{1}{4 \pi \varepsilon_0} \int_V \frac{\rho}{r^2} d V \hat{\mathbf{r}}
\)

Example 12: What charge would be required to electrify a sphere of radius 25 cm, so as to get a surface charge density of \(\frac{3}{\pi} \mathrm{Cm}^{-2}\)?

Solution: Given, \(r=25 \mathrm{~cm}=025 \mathrm{~m}\) and \(\sigma=\frac{3}{\pi} \mathrm{Cm}^{-2}\)
\(\begin{array}{ll}\text { As, } & \sigma=\frac{q}{4 \pi r^2} \\ \therefore & q=4 \pi r^2 \sigma=4 \pi \times(0.25)^2 \times \frac{3}{\pi}=0.75 \mathrm{C}\end{array}\)

Example 13: Sixty-four drops of radius 0.02 m and each carrying a charge of \(5 \mu \mathrm{C}\) are combined to form a bigger drop. Find how the surface charge density of electrification will change, if no charge is lost?

Solution: Volume of each small drop \(=\frac{4}{3} \pi(0.02)^3 \mathrm{~m}^3\)
Volume of 64 small drops \(=\frac{4}{3} \pi(0.02)^3 \times 64 \mathrm{~m}^3\)
Let \(R\) be the radius of the bigger drop formed, then
\(
\frac{4}{3} \pi R^3=\frac{4}{3} \pi(0.02)^3 \times 64
\)
\(
R^3=(0.02)^3 \times 4^3
\)
\(
\therefore \quad R=0.02 \times 4=0.08 \mathrm{~m}
\)
Charge on small drop \(=5 \mu \mathrm{C}=5 \times 10^{-6} \mathrm{C}\)
Surface charge density of small drop,
\(
\sigma_1=\frac{q}{4 \pi r^2}=\frac{5 \times 10^{-6}}{4 \pi(0.02)^2} \mathrm{Cm}^{-2}
\)
Surface charge density of bigger drop,
\(
\sigma_2=\frac{5 \times 10^{-6} \times 64}{4 \pi(0.08)^2} \mathrm{Cm}^{-2}
\)
\(
\frac{\sigma_1}{\sigma_2}=\frac{5 \times 10^{-6}}{4 \pi(0.02)^2} \times \frac{4 \pi(0.08)^2}{5 \times 10^{-6} \times 64}=\frac{1}{4}=1: 4
\)

Electric Field of a Ring of Charge

A conducting ring of radius \(R\) has a total charge \(q\) uniformly distributed over its circumference. We are interested in finding the electric field at point \(P\) that lies on the axis of the ring at a distance \(x\) from its centre.

We divide the ring into infinitesimal segments of length \(d l\). Each segment has a charge \(d q\) and acts as a point charge source of electric field.
Let \(d \mathbf{E}\) be the electric field from one such segment; the net electric field at \(P\) is then the sum of all contributions \(d \mathbf{E}\) from all the segments that make up the ring. If we consider two ring segments at the top and bottom of the ring, we see that the contributions \(d\) E to the field at \(P\) from these segments have the same \(x\)-component but opposite \(y\)-components. Hence, the total \(y\)-component of field due to this pair of segments is zero. When we add up the contributions from all such pairs of segments, the total field \(\mathbf{E}\) will have only a component along the ring’s symmetry axis (the \(x\)-axis) with no component perpendicular to that axis (i.e. no \(y\) or \(z\)-component). So, the field at \(P\) is described completely by its \(x\)-component \(E_x\).
\(\text { Calculation of } E_x\)
\(
\begin{aligned}
d q & =\left(\frac{q}{2 \pi R}\right) \cdot d l \\
d E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{d q}{r^2} \\
d E_x=d E \cos \theta & =\left(\frac{1}{4 \pi \varepsilon_0}\right)\left(\frac{d q}{x^2+R^2}\right)\left(\frac{x}{\sqrt{x^2+R^2}}\right) \\
& =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(d q) x}{\left(x^2+R^2\right)^{3 / 2}}
\end{aligned}
\)
\(
\begin{aligned}
& E_x=\int d E_x=\frac{x}{4 \pi \varepsilon_0\left(x^2+R^2\right)^{3 / 2}} \int d q \\
& E_x=\left(\frac{1}{4 \pi \varepsilon_0}\right) \frac{q x}{\left(x^2+R^2\right)^{3 / 2}}
\end{aligned}
\)

From the above expression, we can see that
(i) \(E_x=0\) at \(x=0\), i.e. field is zero at the centre of the ring. We should expect this, charges on opposite sides of the ring would push in opposite directions on a test charge at the centre, and the forces would add to zero.
(ii) \(E_x=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{x^2}\) for \(x \gg R\), i.e. when the point \(P\) is much farther from the ring, its field is the same as that of a point charge. To an observer far from the ring, the ring would appear like a point, and the electric field reflects this.
(iii) \(E_x\) will be maximum where \(\frac{d E_x}{d x}=0\). Differentiating \(E_x\) w.r.t. \(x\) and putting it equal to zero we get \(x=\frac{R}{\sqrt{2}}\) and \(E_{\max }\) comes out to be, \(\frac{2}{\sqrt[3]{3}}\left(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R^2}\right)\).

Example 14: A charge of \(4 \times 10^{-9} \mathrm{C}\) is distributed uniformly over the circumference of a conducting ring of radius 0.3 m. Calculate the field intensity at a point on the axis of the ring at 0.4 m from its centre and also at the centre.

Solution: Given, \(q=4 \times 10^{-9} \mathrm{C}, x=0.4 \mathrm{~m}\) and \(R=0.3 \mathrm{~m}\)
Electric field intensity at 0.4 m from its centre,
\(
\begin{aligned}
E & =\frac{q x}{4 \pi \varepsilon_0\left(R^2+x^2\right)^{3 / 2}} \\
& =\frac{9 \times 10^9 \times 4 \times 10^{-9} \times 0.4}{\left(03^2+0.4^2\right)^{3 / 2}} \\
E & =\frac{14.4}{(0.5)^3}=115.2 \mathrm{~N} / \mathrm{C}
\end{aligned}
\)
At the centre of the ring, \(x=0\)
\(\therefore\) Electric field intensity, \(E=0\)

Electric Dipole

A pair of equal and opposite point charges \(\pm q\), that are separated by a fixed distance is known as an electric dipole (it is a measure of the separation of positive and negative charges within a system). 

Dipole moment

The product of the magnitude of one charge and the distance between the charges is called the magnitude of the electric dipole moment \(p\). Suppose the charges of the dipole are \(-q\) and \(+q\) and the small distance between them is \(2 a\). Then, the magnitude of the electric dipole moment is given by
\(
{p}=(2 a) q
\)
The electric dipole moment is a vector \(\mathbf{p}\) whose direction is along the line joining the two charges pointing from the negative charge to the positive charge. Its SI unit is Coulomb-metre.

Example 15: \(A\) system has two charges, \(q_A=2.5 \times 10^{-7} \mathrm{C}\) and \(q_B=-2.5 \times 10^{-7} C\) located at points \(A(0,0,-15 \mathrm{~cm})\) and \(B(0,0,+15 \mathrm{~cm})\) respectively. What is the electric dipole moment of the system?

Solution: Electric dipole moment,
\(p=\) magnitude of either charge \(\times\) dipole length
\(
=q_A \times A B=2.5 \times 10^{-7} \times 0.30=7.5 \times 10^{-8} \mathrm{C}-\mathrm{m}
\)

The electric dipole moment is directed from \(B\) to \(A\), i.e. from negative charge to positive charge.

Example 16: Three charges are placed as shown. Find the dipole moment of the arrangements.

Solution: Here, two dipoles are formed as shown in the diagram below

Resultant dipole moment, \(p_r=\sqrt{2} p=\sqrt{2} q d\) and
\(
\theta=45^{\circ}
\)

Example 17: A charge \(q=1 \mu C\) is placed at point ( \(1 \mathrm{~m}, 2 \mathrm{~m}, 4 \mathrm{~m}\) ). Find the electric field at point \(P(0 m,-4 m, 3 m)\).

Solution: Here, \(\quad \mathbf{r}_q=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\)
\(
\mathbf{r}_p=-4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}
\)
\(
\therefore \quad \mathbf{r}_p-\mathbf{r}_q=-\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-\hat{\mathbf{k}}
\)
\(
\left|\mathbf{r}_p-\mathbf{r}_q\right|=\sqrt{(-1)^2+(-6)^2+(-1)^2}=\sqrt{38} \mathrm{~m}
\)
Now, electric field, \(\mathbf{E}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{\left|\mathbf{r}_p-\mathbf{r}_q\right|^3}\left(\mathbf{r}_p-\mathbf{r}_q\right)\)
Substituting the values, we get
\(
\begin{aligned}
& \mathbf{E}=\frac{\left(9.0 \times 10^9\right)\left(1.0 \times 10^{-6}\right)}{(38)^{3 / 2}}(-\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\
& \mathbf{E}=(-38.42 \hat{\mathbf{i}}-230.52 \hat{\mathbf{j}}-38.42 \hat{\mathbf{k}}) \mathrm{N} / \mathrm{C}
\end{aligned}
\)

Dipole field

The electric field produced by an electric dipole is called a dipole field. The total charge of the electric dipole is zero but dipole field is not zero. It can be found using Coulomb’s law and the superposition principle. We will find electric field of an electric dipole at two points as discussed below.

Electric field at an axial point of an electric dipole

Let us calculate electric field at the point \(P\) at a distance \(r\) from the centre of the dipole on the axial line of the dipole on the side of the charge \(q\) as shown in the figure.

\(
\mathbf{E}_{-q}=\frac{-q}{4 \pi \varepsilon_0(r+a)^2}
\)
\(
\mathbf{E}_{+q}=\frac{q}{4 \pi \varepsilon_0(r-a)^2}
\)
\(
\text { The total field at } P \text { is } \mathbf{E}=\mathbf{E}_{+q}+\mathbf{E}_{-q}
\)
\(
\begin{aligned}
& =\frac{q}{4 \pi \varepsilon_0}\left[\frac{1}{(r-a)^2}-\frac{1}{(r+a)^2}\right] \\
& =\frac{q}{4 \pi \varepsilon_0} \frac{4 a r}{\left(r^2-a^2\right)^2} \quad (\because 2 a q=p)
\end{aligned}
\)
\(
\begin{aligned}
\therefore \quad \mathbf{E} & =\frac{2(2 a q) r}{4 \pi \varepsilon_0\left(r^2-a^2\right)^2} \\
& =\frac{2 r \mathbf{p}}{4 \pi \varepsilon_0\left(r^2-a^2\right)^2}
\end{aligned}
\)
\(
=\frac{2 r \mathbf{p}}{4 \pi \varepsilon_0\left(r^2-a^2\right)^2}
\)
For short dipole, i.e. for \(r \gg a, \quad \mathbf{E}=\frac{2 \mathbf{p}}{4 \pi \varepsilon_0 r^3}\) Direction of \(\mathbf{E}\) is same as \(\mathbf{p}\).

Electric field at an equatorial point of an electric dipole

An equatorial point of an electric dipole is any point on the plane that is perpendicular to the dipole’s axis and passes through its center, where the electric potential is zero. Let’s consider an electric dipole whose midpoint is at the origin. The length of the dipole is \(2a\). The magnitude of the charges at the ends is \(q\). So, the magnitude of the dipole moment is, \(p=2 q a\)
The direction of the dipole moment vector is along the axis of the dipole from negative charge to positive charge. Here it is along the positive \(x\)-axis.

We are to find the electric field at a point on the equatorial line (at point \(M\)). Let’s take the point as \(M\) which is at a distance \(r\) away from the midpoint of the dipole i.e. the origin.
The magnitude of the electric field at the point, \(M\) due to the charge \(-q\) is,
\(
\vec{E_{-}}=\frac{1}{4 \pi \epsilon_0} \frac{q}{\left\{\left(r^2+a^2\right)^{\frac{1}{2}}\right\}^2}
\)
The magnitude of the electric field at the point, \(M\) due to the charge \(+q\) is,
\(
\vec{E_{+}}=\frac{1}{4 \pi \epsilon_0} \frac{q}{\left\{\left(x^2+a^2\right)^{\frac{1}{2}}\right\}^2}
\)
The vertical components of \(\mathrm{E}_{+}\) and \(\mathrm{E}_{-}\) cancel each other and the horizontal components add up.
The total electric field \(\vec{E}\) at \(M\) is opposite to dipole moment vector \(\vec{p}\) (from negative to positive). So, we have
\(\vec{E}=-\left[\left(\vec{E_{+}}+\vec{E_{-}}\right) \sin \theta\right]=-2 E \sin \theta \hat{i}, \quad \text { as } \left|\vec{E_{-}}\right|=\left|\vec{E_{+}}\right|=E\) 

\(
=-2 \times \frac{1}{4 \pi \epsilon_0} \frac{q}{\left\{\left(r^2+a^2\right)^{\frac{1}{2}}\right\}^2} \times \frac{a}{\sqrt{r^2+a^2}} \hat{i}=-\frac{1}{4 \pi \epsilon_0} \frac{\vec{p}}{\left(r^2+a^2\right)^{\frac{3}{2}}} \quad [\because \vec{p}=2 a q \hat{i}]
\)
For short dipole, \(r \gg a\)
\(
\vec{E}=\frac{-\vec{p}}{4 \pi \varepsilon_0 r^3}
\)

Note: The electric field due to a short dipole at a large distance \((r \gg\) a) is proportional to \(\frac{1}{r^3}\). If we take the limit, when the dipole size \(2 a\) approaches zero, the charge \(q\) approaches infinity in such a way that the product, \(p=q \times 2 \mathrm{a}\) is finite. Such a dipole is referred to as a point dipole (ideal dipole).

Example 18: Two opposite charges each of magnitude \(2 \mu \mathrm{C}\) are 1 cm apart. Find the electric field at a distance of 5 cm from the mid-point on the axial line of the dipole. Also, find the field on the equatorial line at the same distance from the mid-point.

Solution: Electric field \((E)\) on axial line is given by
\(
\frac{2 p r}{4 \pi \varepsilon_0\left(r^2-a^2\right)^2}
\)
where, \(p\) is dipole moment \(=\) either charge \(\times\) dipole length
Thus, \(p=q \cdot 2 a=\left(2 \times 10^{-6}\right) \times(0.01)\)
Also, \(r=5 \times 10^{-2} \mathrm{~m}\)
\(
\therefore \quad E_a=\frac{9 \times 10^9 \times 2\left(2 \times 10^{-6} \times 10^{-2}\right) \times 5 \times 10^{-2}}{\left[\left(5 \times 10^{-2}\right)^2-\left(0.5 \times 10^{-2}\right)^2\right]^2}
\)
\(
=2.93 \times 10^6 \mathrm{NC}^{-1}
\)
Similarly, electric field \((E)\) on equatorial line is given by
\(
E_e=\frac{p}{4 \pi \varepsilon_0\left(r^2+a^2\right)^{3 / 2}}
\)
The symbols have the same meaning as above,
\(
\begin{array}{ll}
& E_e=\frac{9 \times 10^9 \times\left(2 \times 10^{-6} \times 10^{-2}\right)}{\left[\left(5 \times 10^{-2}\right)^2+\left(0.5 \times 10^{-2}\right)^2\right]^{3 / 2}} \\
\therefore \quad & E_e=1.46 \times 10^6 \mathrm{NC}^{-1}
\end{array}
\)

Electric field at the position \((r, \theta)\)

Due to the positive charge of the dipole, the electric field at point \(P\) will be in radially outward direction, and due to the negative charge, it will be radially inward. Now, we have considered the radial component \(\left(E_r\right)\) and transverse component \(\left(E_\theta\right)\) of the net electric field \((E)\) as shown in the figure.

\(P\) is a general point at Position vector \({r}\) making angle \(\theta\) with \(\vec{p}\). The dipole moment \(\vec{p}\) is resolved in two components one along \({r}(p \cos \theta)\) and one normal to \({r}(p \sin \theta)\)
The electric field at point \(P\) due to the axial line component is
\(
E_r=\frac{2 \vec{p}}{4 \pi \varepsilon_0 r^3}=\frac{1}{4 \pi \varepsilon_0}  \frac{2 p \cos \theta}{r^3}
\)
The electric field at point \(P\) due to the equatorial component is
\(
E_\theta=\frac{-\vec{p}}{4 \pi \varepsilon_0 r^3}=\frac{1}{4 \pi \varepsilon_0}  \frac{p \sin \theta}{r^3}
\)
\(\therefore\) Net electric field at point \(P\) is \(E=\sqrt{E_r^2+E_\theta^2}\)
\(
E=\frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3} \sqrt{1+3 \cos ^2 \theta}
\)
Direction of the electric field, \(\tan \alpha=\frac{E_\theta}{E_r}=\frac{1}{2} \tan \theta\)

Example 19: What is the magnitude of electric field intensity due to a dipole of moment \(2 \times 10^{-8} \mathrm{C}\)-m at a point distance 1 m from the centre of dipole, when line joining the point to the centre of dipole makes an angle of \(60^{\circ}\) with dipole axis?

Solution: Given, \(p=2 \times 10^{-8} \mathrm{C}-\mathrm{m}, r=1 \mathrm{~m}\) and \(\theta=60^{\circ}\)
\(\therefore\) Electric field intensity, \(E=\frac{p}{4 \pi \varepsilon_0 r^3} \sqrt{3 \cos ^2 \theta+1}\)
\(
\begin{aligned}
& =\frac{2 \times 10^{-8} \times 9 \times 10^9}{(1)^3} \times \sqrt{3\left(\cos 60^{\circ}\right)^2+1} \\
& =238.1 \mathrm{~N} / \mathrm{C}
\end{aligned}
\)

Force on dipole

Suppose an electric dipole of dipole moment \(|\mathbf{p}|=2 a q\) is placed in a uniform electric field \(\mathbf{E}\) at an angle \(\theta\), where \(\theta\) is the angle between \(\mathbf{p}\) and \(\mathbf{E}\). A force \(\mathbf{F}_1=q \mathbf{E}\) will act on positive charge and \(\mathbf{F}_2=-q \mathbf{E}\) on negative charge. Since, \(F_1\) and \(F_2\) are equal in magnitude but opposite in direction.

\(
\text { Hence, } \quad \mathbf{F}_1+\mathbf{F}_2=0
\)
\(
\text { or } \quad \mathrm{F}_{\text {net }}=0
\)
Thus, the net force on a dipole in a uniform electric field is zero. While in a non-uniform electric field, it may or may not be zero. The field in which the strength remains the same at all points is known as the uniform electric field. In a uniform electric field, the field strength does not change, and the field lines tend to be parallel and are equally spaced.

Torque on an electric dipole

We know that a dipole, when kept in an external electric field, experiences a rotating effect. The force that causes the rotating effect is known as the torque on the dipole. The two equal and opposite forces shown in the above diagram act at different points of the dipole. They form a couple which exerts a torque. This torque has a magnitude equal to the magnitude of either force multiplied by the arm of the couple, i.e. perpendicular distance between the two anti-parallel forces.

Torque is a vector quantity, Since the force magnitudes are equal and are separated by a distance \(2a\), the torque on the dipole is given by the formula:
Torque \((\tau)=\) Force \(\times\) distance separating forces
Magnitude of torque \(=q E \times 2 a \sin \theta=2 q a E \sin \theta\)
\(
\tau=p E \sin \theta \quad[\because p=(2 a) q]
\)
\(
\mathbf{\tau}=\mathbf{p} \times \mathbf{E}, \text { same as } \vec{\tau}=\vec{p} \times \vec{E}
\)
Thus, the magnitude of torque is \(\tau=p E \sin \theta\). The direction of torque is perpendicular to the plane of paper inwards. Further this torque is zero at \(\theta=0^{\circ}\) or \(\theta=180^{\circ}\), i.e. when the dipole is parallel or anti-parallel to \(\mathbf{E}\) and maximum at \(\theta=90^{\circ}\).
Thus, variation of \(\tau\) with \(\theta\) is as shown in graph below

Example 20: An electric dipole with dipole moment \(4 \times 10^{-9} \mathrm{C}\)-m is aligned at \(30^{\circ}\) with the direction of a uniform electric field of magnitude \(5 \times 10^4 \mathrm{NC}^{-1}\). Calculate the magnitude of the torque acting on the dipole.

Solution: Using the formula, \(\tau=p E \sin \theta \dots(i)\)
Here, dipole moment, \(p=4 \times 10^{-9} \mathrm{C}-\mathrm{m}\) and \(E=5 \times 10^4 \mathrm{NC}^{-1}\)
Angle between \(E\) and \(p, \theta=30^{\circ}\).
Substituting these values in Eq (i), we get
\(
\begin{aligned}
\tau & =4 \times 10^{-9} \times 5 \times 10^4 \times \sin 30^{\circ} \\
& =20 \times 10^{-5} \times \frac{1}{2}=10^{-4} \mathrm{~N}-\mathrm{m}
\end{aligned}
\)

Work done in rotating a dipole in a uniform electric field

When an electric dipole is placed in a uniform electric field \(\mathbf{E}\), [Fig. (a)] a torque, \(\tau=\mathbf{p} \times \mathbf{E}\) acts on it. If we rotate the dipole through a small angle \(d \theta\) as shown in Fig. (b), the work done by the torque is
\(
d W=\tau d \theta \Rightarrow d W=-p E \sin \theta d \theta
\)
The work is negative as the rotation \(d \theta\) is opposite to the torque.

Total work done by external forces in rotating a dipole from \(\theta=\theta_1\) to \(\theta=\theta_2\) [Figs. (c) and (d)] will be given by
\(
W=\int_{\theta_1}^{\theta_2} p E \sin \theta d \theta
\)
\(
W_{\text {extermal force }}=p E\left(\cos \theta_1-\cos \theta_2\right)
\)
and work done by electric forces,
\(
\begin{aligned}
W_{\text {electric force }} & =-W_{\text {external force }} \\
& =p E\left(\cos \theta_2-\cos \theta_1\right)
\end{aligned}
\)
Taking \(\theta_1=\theta\) and \(\theta_2=90^{\circ}\), we have
\(
\begin{aligned}
W_{\text {electric dipole }} & =p \cdot E\left(\cos 90^{\circ}-\cos \theta\right)=-p E \cos \theta \\
& =-\mathbf{p} \cdot \mathbf{E}
\end{aligned}
\)

Important Points:

• If the dipole is placed in non-uniform electric field, then magnitude and direction of electric field is different at every point and it will experience both net force and net torque.
• The work done by external forces to move a charge in an electric field is the negative of the work done by the electric field itself because the external force is acting against the electric force, storing energy as potential energy in the process.

Example 21: An electric dipole of dipole moment \(p=5 \times 10^{-18} C\)-m lying along uniform electric field \(E=4 \times 10^4 N C^{-1}\). Calculate the work done is rotating the dipole by \(60^{\circ}\).

Solution: It is given that, the electric dipole moment,
\(
p=5 \times 10^{-18} \mathrm{C}-\mathrm{m}
\)
Electric field strength, \(E=4 \times 10^4 \mathrm{NC}^{-1}\)
When the electric dipole is placed in an electric field \(\mathbf{E}\), a torque \(\tau=\mathbf{p} \times \mathbf{E}\) acts on it. This torque tries to rotate the dipole through an angle \(\theta\).
If the dipole is rotated from an angle \(\theta_1\) to \(\theta_2\), then work done by an external force is given by
\(
W=p E\left(\cos \theta_1-\cos \theta_2\right) \dots(i)
\)
\(
\begin{aligned}
&\text { Putting } \theta_1=0^{\circ}, \theta_2=60^{\circ} \text { in the Eq. (i), we get }\\
&\begin{aligned}
W & =p E\left(\cos 0^{\circ}-\cos 60^{\circ}\right) \\
& =p E(1-1 / 2)=\frac{p E}{2} \\
& =\frac{5 \times 10^{-18} \times 4 \times 10^4}{2}=10^{-13} \mathrm{~J}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
W & =0.1 \times 10^{-12} \mathrm{~J} \\
& =0.1 \mathrm{pJ}
\end{aligned}
\)

Potential Energy of an Electric Dipole

Potential energy can be associated with the orientation of an electric dipole in an electric field. The dipole has its least potential energy when it is in its equilibrium orientation, which is when its moment \(\vec{p}\) is lined up with the field \(\vec{E}\) (then \(\vec{\tau}=\vec{p} \times \vec{E}=0\) ). It has greater potential energy in all other orientations. Thus the dipole is like a pendulum, which has its least gravitational potential energy in its equilibrium orientation – at its lowest point. To rotate the dipole or the pendulum to any other orientation requires work by some external agent.

In any situation involving potential energy, we are free to define the zero-potential-energy configuration in an arbitrary way because only differences in potential energy have physical meaning. The expression for the potential energy of an electric dipole in an external electric field is simplest if we choose the potential energy to be zero when the angle \(\theta\) in Fig. above is \(90^{\circ}\). We then can find the potential energy \(U\) of the dipole at any other value of \(\theta\) \((\Delta U=-W)\) by calculating the work \(W\) done by the field on the dipole when the dipole is rotated to that value of \(\theta\) from \(90^{\circ}\). We know \(\left(W=\int \tau d \theta\right)\) and \(\tau=-p E \sin \theta\), we find that the potential energy \(U\) at any angle \(\theta\) is
\(
U=-W=-\int_{90^{\circ}}^\theta \tau d \theta=\int_{90^{\circ}}^\theta p E \sin \theta d \theta
\)
\(
\begin{aligned}
&\text { Evaluating the integral leads to }\\
&U=-p E \cos \theta \dots(i)
\end{aligned}
\)
We can generalize this equation to vector form as
\(
U=-\vec{p} \cdot \vec{E} \quad \text { (potential energy of a dipole). } \dots(ii)
\)
Equations (i) and (ii) show us that the potential energy of the dipole is least \((U=-p E)\) when \(\theta=0(\vec{p}\) and \(\vec{E}\) are in the same direction); the potential energy is greatest \((U=p E)\) when \(\theta=180^{\circ}(\vec{p}\) and \(\vec{E}\) are in opposite directions).

When a dipole rotates from an initial orientation \(\theta_i\) to another orientation \(\theta_f\), the work \(W\) done on the dipole by the electric field is
\(
W=-\Delta U=-\left(U_f-U_i\right)
\)
where \(U_f\) and \(U_i\) are calculated with Eq. (ii). If the change in orientation is caused by an applied torque (commonly said to be due to an external agent), then the work \(W_{ext}\) done on the dipole by the applied torque is the negative of the work done on the dipole by the field; that is,
\(
W_a=-W=\left(U_f-U_i\right)
\)

Equilibrium of Dipole

When an electric dipole is placed in a uniform electric field net force on it is zero for any position of the dipole in the electric field. But torque acting on it is zero only at \(\theta=0^{\circ}\) and \(180^{\circ}\). Thus, we can say that at these two positions of the dipole, net force and torque on it is zero or the dipole is in equilibrium.

Of this, \(\theta=0^{\circ}\) is the stable equilibrium position of the dipole because potential energy in this position is minimum \(\left(U=-p E \cos 0^{\circ}=-p E\right)\) and when displaced from this position a torque starts acting on it which is restoring in nature and which has a tendency to bring the dipole back in its equilibrium position. On the other hand, at \(\theta=180^{\circ}\), the potential energy of the dipole is maximum \(\left(U=-p E \cos 180^{\circ}=+p E\right)\) and when it is displaced from this position, the torque has a tendency to rotate it in other direction. This torque is not restoring in nature. So, this equilibrium is known as unstable equilibrium position.

Example 22: Draw electric lines of forces due to an electric dipole. 

Solution: Electric lines of forces due to an electric dipole are as shown in the figure. The electric lines of force due to an electric dipole originate from the positive charge and terminate at the negative charge, forming a characteristic “bow-tie” shape, because the electric field is stronger near the charges and weaker further away.

Note: In the above figure, we can see that direction of the electric field is in the opposite direction of \(\mathbf{p}\) (starts at negative charge and ends at positive charge) between the two charges.

Example 23: At a far away distance \(r\) along the axis from an electric dipole electric field is \(E\). Find the electric field at distance \(2 r\) along the perpendicular bisector.

Solution: Along the axis of the dipole,
\(
E=\frac{1}{4 \pi \varepsilon_0} \frac{2 p}{r^3} \dots(i)
\)
This electric field is in the direction of \(\mathbf{p}\) Along the perpendicular bisector at a distance \(2 r\),
\(
E^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{p}{(2 r)^3} \dots(ii)
\)
From Eqs. (i) and (ii), we can see that
\(
E^{\prime}=\frac{E}{16}
\)
Moreover, \(E^{\prime}\) is in the opposite direction of \(\mathbf{p}\) Hence,
\(
\mathbf{E}^{\prime}=-\frac{\mathbf{E}}{16}
\)

Example 24: A neutral water molecule \(\left(\mathrm{H}_2 \mathrm{O}\right)\) in its vapor state has an electric dipole moment of magnitude \(6.2 \times 10^{-30} \mathrm{C} \cdot \mathrm{m}\).
(a) How far apart are the molecule’s centers of positive and negative charge?
(b) If the molecule is placed in an electric field of \(1.5 \times\) \(10^4 \mathrm{~N} / \mathrm{C}\), what maximum torque can the field exert on it? (Such a field can easily be set up in the laboratory.)
(c) How much work must an external agent do to rotate this molecule by \(180^{\circ}\) in this field, starting from its fully aligned position, for which \(\theta=0\)?

Solution:(a) There are 10 electrons and 10 protons in a neutral water molecule; so the magnitude of its dipole moment is
\(
p=q d=(10 e)(d)
\)
in which \(d\) is the separation we are seeking and \(e\) is the elementary charge. Thus,
\(
\begin{aligned}
d & =\frac{p}{10 e}=\frac{6.2 \times 10^{-30} \mathrm{C} \cdot \mathrm{~m}}{(10)\left(1.60 \times 10^{-19} \mathrm{C}\right)} \\
& =3.9 \times 10^{-12} \mathrm{~m}=3.9 \mathrm{pm}
\end{aligned}
\)
This distance is not only small, but it is also actually smaller than the radius of a hydrogen atom.

(b) The torque on a dipole is maximum when the angle \(\theta\) between \(\vec{p}\) and \(\vec{E}\) is \(90^{\circ}\).
Substituting \(\theta=90^{\circ}\)
\(
\begin{aligned}
\tau & =p E \sin \theta \\
& =\left(6.2 \times 10^{-30} \mathrm{C} \cdot \mathrm{~m}\right)\left(1.5 \times 10^4 \mathrm{~N} / \mathrm{C}\right)\left(\sin 90^{\circ}\right) \\
& =9.3 \times 10^{-26} \mathrm{~N} \cdot \mathrm{~m} .
\end{aligned}
\)

(c) The work done by an external agent (by means of a torque applied to the molecule) is equal to the change in the molecule’s potential energy due to the change in orientation.
\(
\begin{aligned}
W_a & =U_{180^{\circ}}-U_0 \\
& =\left(-p E \cos 180^{\circ}\right)-(-p E \cos 0) \\
& =2 p E=(2)\left(6.2 \times 10^{-30} \mathrm{C} \cdot \mathrm{~m}\right)\left(1.5 \times 10^4 \mathrm{~N} / \mathrm{C}\right) \\
& =1.9 \times 10^{-25} \mathrm{~J} .
\end{aligned}
\)

Area Vector

For discussing the flux of a vector field, it is helpful to introduce an area vector \(\overrightarrow{\mathbf{A}}\). This allows us to write the last equation in a more compact form. What should the magnitude of the area vector be?

The area vector of a flat surface of area \(A\) has the following magnitude and direction:
Magnitude is equal to area \((A)\)
Direction is along the normal to the surface ( \((\widehat{\mathbf{n}})\); that is, perpendicular to the surface.

Since the normal to a flat surface can point in either direction from the surface, the direction of the area vector of an open surface needs to be chosen, as shown in the Figure below:

Since \(\widehat{\mathbf{n}}\) is a unit normal to a surface, it has two possible directions at every point on that surface (figure(a)). For an open surface, we can use either direction, as long as we are consistent over the entire surface. However, if a surface is closed, then the surface encloses a volume. In that case, the direction of the normal vector at any point on the surface points from the inside to the outside. On a closed surface such as that of Figure (b), \(\widehat{\mathbf{n}}\) is chosen to be the outward normal at every point, to be consistent with the sign convention for electric charge.

Electric Flux

Electric flux over an area in an electric field is a measure of the number of field lines crossing a surface. Electric flux is a scalar quantity having SI unit V-m or \(\mathrm{N}-\mathrm{m}^2 \mathrm{C}^{-1}\)

Flat Surface, Uniform Field. We begin with a flat surface with area \(S\) in a uniform electric field \(\vec{E}\). Figure (a) shows one of the electric field vectors \(\vec{E}\) piercing a small square patch with area \(\Delta S\) (where \(\Delta\) indicates “small”). Actually, only the \(x\) component (with magnitude \(E_x=E \cos \theta\) in Fig. (b)) pierces the patch. The \(y\) component merely skims along the surface (no piercing in that) and does not come into play in Gauss’ law. The amount of electric field piercing the patch is defined to be the electric flux \(\boldsymbol{\Delta} \Phi\) through it:
\(
\Delta \Phi=(E \cos \theta) \Delta S
\)
We can write the above equation in vector form as
\(
\Delta \Phi=\vec{E} \cdot \Delta \vec{S}
\)
and the dot product automatically gives us the component of \(\vec{E}\) that is parallel to \(\Delta \vec{S}\) and thus piercing the patch.

To find the total flux \(\Phi\) through the surface, we sum the flux through every patch on the surface:
\(
\Phi=\sum \vec{E} \cdot \Delta \vec{S}
\)
However, because we do not want to sum hundreds (or more) flux values, we transform the summation into an integral by shrinking the patches from small squares with area \(\Delta S\) to patch elements (or area elements) with area \(d S\). The total flux is then
\(
\Phi=\int \vec{E} \cdot d \vec{S} \quad \text { (total flux) }
\)
Now we can find the total flux by integrating the dot product over the full surface.

Dot Product: (The flux of the electric field passing through an area is the dot product of electric field vector and area vector)
When the electric field is uniform and the surface is flat, the product \(E \cos \theta\) is a constant and comes outside the integral. The remaining \(\int d S\) is just an instruction to sum the areas of all the patch elements to get the total area, but we already know that the total area is \(S\). So the total flux in this simple situation is
\(
\Phi=\int \vec{E} \cdot d \vec{S}=(E \cos \theta) \int d \vec{S} =(E \cos \theta) S = (ES \cos \theta) \quad \text { (uniform field, flat surface). }
\)

How does Field pierce in a Closed Surface?

The figure below shows the electric field of an oppositely charged, parallel-plate system and an imaginary box between the plates. The electric field between the plates is uniform and points from the positive plate toward the negative plate. A calculation of the flux of this field through various faces of the box shows that the net flux through the box is zero. Why does the flux cancel out here?

The reason is that the sources of the electric field are outside the box. Therefore, if any electric field line enters the volume of the box, it must also exit somewhere on the surface because there is no charge inside for the lines to land on. Therefore, quite generally, electric flux through a closed surface is zero if there are no sources of electric field, whether positive or negative charges, inside the enclosed volume. In general, when field lines leave (or “flow out of”) a closed surface, \(\Phi\) is positive; when they enter (or “flow into”) the surface, \(\Phi\) is negative.

Directions: To keep track of the piercing direction, we again use an area vector \(\Delta \vec{S}\) that is perpendicular to a patch, but now we always draw it pointing outward from the surface (away from the interior). Then if a field vector pierces outward, it and the area vector are in the same direction, the angle is \(\theta=0\), and \(\cos \theta=1\). Thus, the dot product \(\vec{E} \cdot \Delta \vec{S}\) is positive and so is the flux.

Conversely, if a field vector pierces inward, the angle is \(\theta=180^{\circ}\) and \(\cos \theta=-1\). Thus, the dot product is negative and so is the flux. If a field vector skims the surface (no piercing), the dot product is zero (because \(\cos 90^{\circ}=0\) ) and so is the flux.

In principle, to find the net flux through the surface shown above, we find the flux at every patch and then sum the results (with the algebraic signs included). However, we are not about to do that much work. Instead, we shrink the squares to patch elements with area vectors \(d \vec{S}\) and then integrate:
\(
\Phi=\oint \vec{E} \cdot d \vec{S} \quad \text { (net flux) }
\)

Example 25: The electric field in a region is given by \(\mathbf{E}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}\), here, \(a\) and \(b\) are constants. Find the net flux passing through a square area of side \(l\) parallel to \(y z\)-plane.

Solution: A square area of side \(l\) parallel to \(y z\)-plane in vector form can be written as \(\quad \mathbf{S}=l^2 \hat{\mathbf{i}}\)
Given, \(\mathbf{E}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}\)
\(\therefore\) Electric flux passing through the given area will be
\(
\begin{aligned}
\phi & =\mathbf{E} \cdot \mathbf{S} \\
& =(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}) \cdot\left(l^2 \hat{\mathbf{i}}\right)=a l^2
\end{aligned}
\)

Example 26: A rectangular surface of sides 10 cm and 15 cm is placed inside a uniform electric field of \(25 \mathrm{NC}^{-1}\), such that normal to the surface makes an angle of \(60^{\circ}\) with the direction of the electric field. Find the flux of the electric field through the rectangular surface.

Solution: Here, \(E=\) electric field \(=25 \mathrm{NC}^{-1}\)
\(S=\) surface area of rectangle
\(
=l \times b=0.10 \times 0.15 \mathrm{~m}^2
\)
Flux, \(\phi=E S \cos \theta=(25)(0.15 \times 0.10)\left(\cos 60^{\circ}\right)\)
\(
=0.1875 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1}
\)

Example 27: The electric field in a region is given by \(\mathbf{E}=\frac{E_0}{a} x \hat{\mathbf{i}}\). Find the electric flux passing through a cubical volume bounded by the surfaces \(x=0, x=a, y=0, y=a\), \(z=0\) and \(z=a\).

Solution: The given situation is shown below

On four faces, the electric field and area vector are perpendicular, hence there will be no flux. One face is at origin, i.e. \(x=0\) \(\Rightarrow E=0\), hence there will be no flux. On the sixth face, \(x=a\)

\(
\therefore \text { Net electric flux, } \phi=\mathbf{E} \cdot \mathbf{S}=E_0 \hat{\mathbf{i}} \cdot a^2 \hat{\mathbf{i}}=E_0 a^2
\)

Example 28: A cylinder is placed in a uniform electric field \(\mathbf{E}\) with axis parallel to the field. Find the total electric flux through the cylinder.

Solution: The direction of \(\mathbf{E}\) and \(d \mathbf{S}\) on different sections of the cylinder are shown below.

\(
\phi=\int_{\text {Left plane face }} \mathbf{E} \cdot d \mathbf{S}+\int_{\text {Right plane face }} \mathbf{E} \cdot d \mathbf{S}+\int_{\text {curved surface }} \mathbf{E} \cdot d \mathbf{S}
\)
\(
\begin{aligned}
& =\int \mathbf{E} \cdot d \mathbf{S} \cos 180^{\circ}+\int \mathbf{E} \cdot d \mathbf{S} \cos 0^{\circ}+\int \mathbf{E} \cdot d \mathbf{S} \cos 90^{\circ} \\
& =-\mathbf{E} \int d \mathbf{S}+\mathbf{E} \int d \mathbf{S}+\mathbf{0} \\
& =-\mathbf{E} \times \pi r^2+\mathbf{E} \times \pi r^2=0
\end{aligned}
\)

Gauss’ Law

According to Gauss’s law, “the net electric flux through any closed surface is equal to the net charge enclosed by it divided by \(\varepsilon_0 “\). Mathematically, it can be
written as
\(
\Phi= \frac{q_{\text {enc }}}{\varepsilon_0} = \oint \vec{E} \cdot d \vec{s} \quad \text { (Gauss’ law). } \dots(i)
\)
The above equation holds only when the net charge is located in a vacuum (or in air). In Eqs. (i), the net charge \(q_{\text {enc }}\) is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero. We include the sign, rather than just use the magnitude of the enclosed charge, because the sign tells us something about the net flux through the Gaussian surface: If \(q_{\mathrm{enc}}\) is positive, the net flux is outward; if \(q_{\text {enc }}\) is negative, the net flux is inward.

Let us apply these ideas to the Figure above, which shows two particles, with charges equal in magnitude but opposite in sign, and the field lines describing the electric fields the particles set up in the surrounding space. Four Gaussian surfaces (the surface that we choose for the application of Gauss’s law is called the Gaussian surface) are also shown, in cross-section. Let us consider each in turn.

Surface \(\mathbf{S}_1\): The electric field is outward for all points on this surface. Thus, the flux of the electric field through this surface is positive, and so is the net charge within the surface, as Gauss’ law requires. (That is, in Eq. (i), if \(\Phi\) is positive, \(q_{\text {enc }}\) must be also.)
Surface \(\mathbf{S}_2\): The electric field is inward for all points on this surface. Thus, the flux of the electric field through this surface is negative and so is the enclosed charge, as Gauss’ law requires.
Surface \(\mathbf{S}_3\): This surface encloses no charge, and thus \(q_{\text {enc }}=0\). Gauss’ law (Eq. (i)) requires that the net flux of the electric field through this surface be zero. That is reasonable because all the field lines pass entirely through the surface, entering it at the top and leaving at the bottom.
Surface \(\mathbf{S}_4\): This surface encloses no net charge because the enclosed positive and negative charges have equal magnitudes. Gauss’ law requires that the net flux of the electric field through this surface be zero. That is reasonable because there are as many field lines leaving surface \(S_4\) as entering it.

Simplified Form of Gauss’s Theorem

Gauss’s law in simplified form  (say \(E\) is constant) can be written as under
\(
\phi=\oint \vec{E} \cdot d \vec{s}= E \oint d \vec{s}=E S=\frac{q_{\text {enc }}}{\varepsilon_0} \quad \text { or } \quad E=\frac{q_{\text {enc }}}{S \varepsilon_0}
\)
but this form of Gauss’s law is applicable only under the following two conditions :
(i) The electric field at every point on the surface is either perpendicular or tangential.
(ii) The magnitude of the electric field at every point where it is perpendicular to the surface has a constant value (say \(E\) ).
Here, \(S\) is the area where the electric field is perpendicular to the surface.

Special Cases:

Case-1: If the surface contains a number of charges, as shown in the figure, then \(q_{\text {enc }}\) can be calculated as

\(
\begin{aligned}
q_{\mathrm{enc}} & =q_1-q_2+q_3 \\
\phi & =\frac{q_{\text {enc }}}{\varepsilon_0}=\frac{q_1-q_2+q_3}{\varepsilon_0}
\end{aligned}
\)

Case-2: If a charge \(q\) is placed at the centre of a cube, then the flux passing through cube,
\(
\phi=\frac{q_{\text {enc }}}{\varepsilon_0}=\frac{q}{\varepsilon_0}
\)
The flux passing through each face of cube,
\(
\phi^{\prime}=\frac{\phi}{6}=\frac{q}{6 \varepsilon_0} \text { (by symmetry) }
\)

Case-3: To calculate flux passing through an open surface, first make the surface close in such a manner that the point charge comes at the centre and then apply the symmetry concept, e.g.

(a) A charge \(q\) is placed at the centre of an imaginary hemispherical surface.

First, make the surface close by placing another hemisphere.

Flux through sphere, \(\phi=\frac{q_{\text {enc }}}{\varepsilon_0}=\frac{q}{\varepsilon_0}\)
By symmetry, flux through hemisphere, \(\phi^{\prime}=\frac{\phi}{2}=\frac{q}{2 \varepsilon_0}\)

(b) A charge \(Q\) is placed at a distance \(a / 2\) above the centre of a horizontal square of the edge \(a\).

First, make the surface close by placing five square faces \((a \times a)\), so that a cube is formed and charge \(Q\) is at centre of a cube.

Flux through the cube, \(\phi=\frac{q_{\text {enc }}}{\varepsilon_0}=\frac{Q}{\varepsilon_0}\)
By symmetry, flux through each square face,
\(
\phi^{\prime}=\frac{\phi}{6}=\frac{Q}{6 \varepsilon_0}
\)

Note: In the case of a closed symmetrical body with charge \(q\) at its centre, the electric flux linked with each half will be \(\frac{\phi}{2}=\frac{q}{2 \varepsilon_0}\). If the symmetrical closed body has \(n\) identical faces with point charge at its centre, flux linked with each face will be \(\frac{\phi}{n}=\frac{q}{n \varepsilon_0}\).

Example 29: An uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of \(80.0 \mu \mathrm{C} / \mathrm{m}^2\).
(i) Find the charge on the sphere. (ii) What is the total electric flux leaving the surface of the sphere?

Solution: (i) Using the relation \(\sigma \text { (charge per unit area) } =\frac{q}{4 \pi R^2}\), we get
\(
\begin{aligned}
q & =4 \pi R^2 \times \sigma=4 \times \frac{22}{7} \times(1.2)^2 \times 80 \times 10^{-6} \\
& =1.45 \times 10^{-3} \mathrm{C}
\end{aligned}
\)
(ii) Using Gauss’s theorem, we get
\(
\begin{aligned}
\phi & =\frac{q}{\varepsilon_0}=\frac{1.45 \times 10^{-3}}{8.854 \times 10^{-12}} \\
& =1.64 \times 10^8 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1}
\end{aligned}
\)

Example 30: A point charge causes an electric flux of \(-1.0 \times 10^3 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1}\) to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (i) If the radius of the Gaussian surface is doubled, how much flux will pass through the surface? (ii) What is the value of the point charge?

Solution: (i) According to Gauss’s law, the electric flux through a Gaussian surface depends upon the charge enclosed inside the surface and not upon its size (radius). Thus, the electric flux will remain unchanged,
\(
\text { i.e. }-1.0 \times 10^3 \mathrm{~N}-\mathrm{m}^2 \mathrm{C}^{-1}
\)
(ii) Using the formula \(\phi=\frac{q}{\varepsilon_0}\) (Gauss’s theorem), we get
\(
\begin{aligned}
\Rightarrow \quad q & =\varepsilon_0 \phi=8.854 \times 10^{-12} \times\left(-1.0 \times 10^3\right) \\
& =-8.854 \times 10^{-9} \mathrm{C}=-8.8 \mathrm{nC}
\end{aligned}
\)

Example 31: A point charge \(q\) is placed at the centre of a cube. What is the flux linked
(i) with all the faces of the cube?
(ii) with each face of the cube?
(iii) if the charge is not at the centre, then what will be the answers of parts (i) and (ii)?

Solution: (i) According to Gauss’s law, \(\phi_{\text {total }}=\frac{q_{\text {enc }}}{\varepsilon_0}=\frac{q}{\varepsilon_0}\)
(ii) The cube is a symmetrical body with 6 faces and the point charge is at its centre, so electric flux linked with each face will be
\(
\phi_{\text {each face }}=\frac{\phi_{\text {total }}}{6}=\frac{q}{6 \varepsilon_0}
\)

(iii) If the charge is not at the centre, the answer of the part (i) will remain the same while that of the part (ii) will change.

Example 32: A point charge \(Q\) is placed at one corner of a cube. Find flux passing through a cube.

Solution: First, make the surface close by placing three identical cubes at three sides of the given cube and four cubes above. Now, the charge comes at the centre of 8 cubes.
The flux passing through each cube will be (1/8)th of the flux \(Q / \varepsilon_0\). Hence, flux passing through the given cube is \(Q / 8 \varepsilon_0\). Flus through each face of the cube is \(\Phi_{\text {each face }}=\frac{\Phi_{\text {cube }}}{3}=\frac{q}{8 \epsilon_0 \cdot 3}=\frac{q}{24 \epsilon_0}\)

Example 33: A hemispherical body of radius \(R\) is placed in a uniform electric field \(E\). What is the flux linked with the curved surface, if the field is
(i) parallel to the base
(ii) and perpendicular to the base?

Solution: We know, flux passing through closed surface,
\(
\phi=\oint \mathbf{E} \cdot d \mathbf{S}=\frac{q_{\mathrm{enc}}}{\varepsilon_0}
\)

(i)

\(\begin{aligned} & \text { Charge inside hemisphere, } q_{\text {in }}=0 \text {, } \\ & \text { i.e. } \quad \oint \mathbf{E} \cdot d \mathbf{S}=0\end{aligned}\)
\(
\begin{aligned}
\phi_{\text {curved }}+\phi_{\text {plane }} & =0 \\
\phi_{\text {curved }}+E S \cos 90^{\circ} & =0 \\
\phi_{\text {curved }} & =0
\end{aligned}
\)

(ii) 

Also,\(\quad \phi_{\text {curved }}+\phi_{\text {plane }}=0\)
\(\begin{aligned} \phi_{\text {curved }}+E S \cos 0^{\circ} & =0 \\ \phi_{\text {curved }}+E \pi R^2 & =0 \\ \phi_{\text {curved }} & =-E \pi R^{2}\end{aligned}\)

Applications of Gauss’s law

To calculate the electric field by Gauss’s theorem, we will draw a Gaussian surface (either sphere or cylinder, according to the situation) in such a way that the electric field is perpendicular at each point of the surface and its magnitude is the same at every point and then apply Gauss’s law. 

Electric field due to a point charge

The electric field due to a point charge is everywhere radial. We wish to find the electric field at a distance \(r\) from the charge \(q\). We select Gaussian surface, a sphere at distance \(r\) from the charge. At every point of this sphere, the electric field has the same magnitude \(E\) and it is perpendicular to the surface itself. Hence, we can apply the simplified form of Gauss’s law,
\(
E S=\frac{q_{\text {enc }}}{\varepsilon_0}
\)
Here, \(S=\) area of sphere \(=4 \pi r^2\) and \(q_{\text {enc }}=\) net charge enclosing the Gaussian surface \(=q\)
\(
\begin{aligned}
E\left(4 \pi r^2\right) & =\frac{q}{\varepsilon_0} \\
E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}
\end{aligned}
\)
It is nothing but Coulomb’s law.

Electric field due to an infinitely long straight uniformly charged wire

Consider a long line charge with a linear charge density (charge per unit length), \(\lambda\). To calculate the electric field at a point, located at a distance \(r\) from the line charge, we construct a Gaussian surface, a cylinder of any arbitrary length \(l\) of radius \(r\) and its axis coinciding with the axis of the line charge. This cylinder has three surfaces. One is curved surface and the two plane parallel surfaces.

Field lines at plane parallel surfaces are tangential, so flux passing through these surfaces is zero. The magnitude of electric field is having the same magnitude (say \(E\) ) at curved surface and simultaneously the electric field is perpendicular at every point of this surface.

Hence, we can apply the Gauss’s law as
\(
E S=\frac{q_{\mathrm{in}}}{\varepsilon_0}
\)
Here, \(\quad S=\) area of curved surface \(=(2 \pi r l)\) and \(\quad q_{\text {enc }}=\) net charge enclosing this cylinder \(=\lambda l\).
\(\therefore \quad E(2 \pi r l)=\frac{\lambda l}{\varepsilon_0}\)
\(\therefore\) \(E=\frac{\lambda}{2 \pi \varepsilon_0 r}\)
\(E \propto \frac{1}{r}\)
So, \(E-r\) graph is a rectangular hyperbola as shown in the figure.

Electric field due to a finite length of straight charged wire

Consider a long finite line charge with a linear charge density \(\lambda\)(charge per unit length). We have to calculate the electric field at a point (\(P\)), located at a distance \(r\) from the line charge. Let’s take a small element on the line charge and call it \(dy\).
So the small charge (\(dq\)) on this small element \(dy\) will be \(=\lambda d y\).
The electric field at point \(P\) due to this small charge \(dq\) is \(dE\) which has two components \(dE_x\) in the \(x\) direction and \(dE_y\) in the \(y\) direction.
Therefore, the electric field due to the point charge at point \(P\) is given by
\(
d E=\frac{k(d q)}{\left(\sqrt{r^2+y^2}\right)^2} \quad [k=\frac{1}{4 \pi \varepsilon_0}]
\)
As we move upward from \(dy\) to the top of the wire \(\theta\) increases and let’s take that angle as \(\theta_{1}\) and similarly as we move to the bottom of the wire let’s assume that angle is \(\theta_{2}\).
Now, we can write (we need to bring all parameters in terms of angle)
\(
\begin{aligned}
& \tan \theta=\frac{y}{r} \\
& y=r \tan \theta
\end{aligned}
\)
\(
d y=r \sec ^2 \theta d \theta
\)
\(
E_x=\int_{-\theta_2}^{\theta_1} d E_x  =\int_{-\theta_2}^{\theta_1}  \frac{k(\lambda dy )}{\left(r^2+y^2\right)}
\)
To calculate the electric field at point \(P\) of the finite wire, we have to integrate \(dE\).
First we have to calculate the \(x\)-component of the electric field \(dE_x\) and then \(y\)-component of the electric field \(dE_y\)
\(
E_x=\int_{-\theta_2}^{\theta_1} \frac{k \lambda r \sec ^2 \theta \cos \theta d \theta} {\left(r^2+r^2 \tan ^2 \theta\right)}
\)

\(
=\int_{-\theta_2}^{\theta_1} \frac{k \lambda r \sec ^2 \theta d \theta \cos \theta}{r^2\left(1+\tan ^2 \theta\right)}
\)
\(
=\frac{k \lambda}{r}[\sin \theta_1+\sin \theta_2]
\)
\(
=\frac{1}{4 \pi \varepsilon_0} \frac{\lambda}{r}\left(\sin \theta_1+\sin \theta_2\right)
\)
Similarly, we can calculate \(E_y\)
\(E_y=\int_{-\theta_2}^{\theta_1} d E_y = \int_{-\theta_2}^{\theta_1} \frac{k \lambda r \sec ^2 \theta d \theta \sin \theta}{r^2(1+\tan \delta)}\)
\(
E_y=\frac{k \lambda}{r}[\cos \theta_2-\cos \theta_1)
\)
Now total electric field at point \(P\) will be
\(
\mathbf{E}=E_x \hat{\mathbf{i}}+E_y \hat{\mathbf{(-j)}}
\)

Note: If we consider infinite wire then \(\theta_1=\theta_2=90^{\circ}\)
\(E_x=\frac{1}{4 \pi \varepsilon_0} \frac{2\lambda}{r}\) and \(E_y=0\) 
\(
E=\frac{\lambda}{2 \pi \varepsilon_0 r}
\)

Example 34: An infinite line charge produces a field of \(9 \times 10^4 \mathrm{NC}^{-1}\) at a distance of 2 cm. Calculate the linear charge density.

Solution: As we know that, electric field due to an infinite line charge is given by the relation, \(E=\frac{1}{2 \pi \varepsilon_0} \cdot \frac{\lambda}{r}\)
where \(\lambda\) is linear charge density and \(r\) is the distance of a point where an electric field is produced from the line charge.
\(
\lambda=2 \pi \varepsilon_0 r E
\)
Here, \(E=9 \times 10^4 \mathrm{NC}^{-1}\) and \(r=2 \mathrm{~cm}=0.02 \mathrm{~m}\)
\(\therefore\) Linear charge density,
\(
\lambda=\frac{1}{9 \times 10^9} \times \frac{0.02 \times 9 \times 10^4}{2}=10^{-7} \mathrm{Cm}^{-1}
\)

Example 35: A long cylindrical wire carries a positive charge of linear density \(\lambda\). An electron \((-e, m)\) revolves around it in a circular path under the influence of the attractive electrostatic force. Find the speed of the electron.

Solution:

Electric field at perpendicular distance \(r\),
\(
E=\frac{\lambda}{2 \pi \varepsilon_0 r}
\)
The electric force on electron, \(F=e E\) To move in a circular path, necessary centripetal force is provided by the electric force.
\(
F_c=\frac{m v^2}{r} \Rightarrow \frac{m v^2}{r}=e E=\frac{e \lambda}{2 \pi \varepsilon_0 r} \Rightarrow v=\sqrt{\frac{e \lambda}{2 \pi \varepsilon_0 m}}
\)
\(
\therefore \text { Speed of the electron, } \quad v=\sqrt{\frac{e \lambda}{2 \pi \varepsilon_0 m}}
\)

Field due to a uniformly charged infinite plane sheet

Consider a flat thin sheet, infinite in size with constant surface charge density \(\sigma\) (charge per unit area).

Let us draw a Gaussian surface (a cylinder) with one end on one side and other end on the other side and of crosssectional area \(S_0\). Field lines will be tangential to the curved surface, so flux passing through this surface is zero. At plane surfaces, electric field has same magnitude and perpendicular to surface. Hence, using
\(
\begin{aligned}
E S & =\frac{q_{\text {enc }}}{\varepsilon_0} \\
E\left(2 S_0\right) & =\frac{(\boldsymbol{\sigma})\left(S_0\right)}{\varepsilon_0} \\
E & =\frac{\sigma}{2 \varepsilon_0}
\end{aligned}
\)
Vectorically, \(\mathbf{E}=\frac{\sigma}{2 \varepsilon_0} \hat{\mathbf{n}}\)
Where \(\hat{\mathbf{n}}\) is a unit vector normal to the plane and going away from it.

Example 36: A large plane sheet of charge having surface charge density \(5 \times 10^{-6} \mathrm{Cm}^{-2}\) lies in the XY-plane. Find the electric flux through a circular area of radius 0.1 m, if the normal to the circular area makes an angle of \(60^{\circ}\) with the Z-axis.

Solution: Given, \(\sigma=5 \times 10^{-6} \mathrm{Cm}^{-2}, r=0.1 \mathrm{~m}\) and \(\theta=60^{\circ}\)
\(
\text { Flux, } \quad \phi=E S \cos \theta=\left(\frac{\sigma}{2 \varepsilon_0}\right) \pi r^2 \cos \theta
\)
\(
\begin{aligned}
& =\frac{5 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \times \frac{22}{7} \times(0.1)^2 \cos 60^{\circ} \\
& \phi=4.44 \times 10^3 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}
\end{aligned}
\)

Example 37: Two large, thin metal plates are placed parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude \(1.77 \times 10^{-11}\) coulomb per square metre. What is the electric field
(i) to the left of the plates,
(ii) to the right of the plates
(iii) and in between the plates?

Solution: (i) Electric fields due to both the plates outside them, will be equal in magnitude and opposite in direction, so the net field will be zero.
(ii) The electric field outside the plates will be equal in magnitude and opposite in direction so the net electric field will be zero.
(iii) In between the plates, the electric fields due to both the plates will be adding up, so the net field will be
\(\frac{\sigma}{2 \varepsilon_0}+\frac{\sigma}{2 \varepsilon_0}=\frac{\sigma}{\varepsilon_0}\) from positive to negative plate.
\(\therefore \quad E=\frac{\sigma}{\varepsilon_0}=\frac{1.77 \times 10^{-11}}{8.85 \times 10^{-12}}=2 \mathrm{~N} / \mathrm{C}\)

Electric field near a charged conducting surface

When a charge is given to a conducting plate, it distributes itself over the entire outer surface of the plate. The surface charge density \(\sigma\) is uniform and is the same on both surfaces, if the plate is of uniform thickness and of infinite size. This is similar to the previous one, the only difference is that, this time charges are on both sides. Hence, applying, \(E S=\frac{q_{\text {enc }}}{\varepsilon_0}\)

\(
\begin{aligned}
&\begin{array}{lc}
\text { Here, } & S=2 S_0 \text { and } q_{\text {enc }}=(\sigma)\left(2 S_0\right) \\
\therefore & E\left(2 S_0\right)=\frac{(\sigma)\left(2 S_0\right)}{\varepsilon_0} \\
\therefore & E=\frac{\sigma}{\varepsilon_0}
\end{array}\\
&\text { Thus, the field due to a charged conducting plate is twice the field due to plane sheet of charge. }
\end{aligned}
\)

Example 38: An electric dipole is placed at the centre of a sphere. Find the electric flux passing through the sphere.

Solution:

Net charge inside the sphere \(q_{\text {enc }}=0\). Therefore, according to Gauss’s law net flux passing through the sphere is zero.

Electric field due to a uniformly charged thin spherical shell

Let \(O\) be the centre and \(R\) be the radius of a thin, isolated spherical shell or solid conducting sphere carrying a charge \(+q\) which is uniformly distributed on the surface. We have to determine electric field intensity due to this shell at points outside the shell, on the surface of the shell, and inside the shell.

At external point

We can construct a Gaussian surface (a sphere) of radius \(r>R\). At all points of this sphere, the magnitude of the electric field is the same and its direction is perpendicular to the surface. Thus, we can apply Gauss’s theorem,
\(
\begin{gathered}
E S=\frac{q_{\text {enc }}}{\varepsilon_0} \text { or } E\left(4 \pi r^2\right)=\frac{q}{\varepsilon_0} \\
E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}
\end{gathered}
\)
Hence, the electric field at any external point is the same as, if the total charge is concentrated at centre.
At the surface of the sphere, \(r=R\)
\(
\therefore \quad E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R^2}
\)

At an internal point

In this case, the Gaussian surface encloses no charge, i.e.
\(
\phi=E\left(4 \pi r^2\right)=0, E_{\text {inside }}=0
\)
The electric field intensity is zero everywhere inside the charged shell.
The variation of the electric field \((E)\) with the distance from the centre \((r)\) is as shown in the figure.

Note:
(i) At the surface, the graph is discontinuous.
(ii) \(E_{\text {surface }}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R^2}=\frac{q / 4 \pi R^2}{\varepsilon_0}=\frac{\sigma}{\varepsilon_0}\)

Example 39: A thin spherical shell of metal has a radius of 0.25 m and carries the charge of \(0.2 \mu \mathrm{C}\). Calculate the electric intensity at 3.0 m from the centre of the shell.

Solution: The intensity at an external point at a distance \(r\) from the centre of the shell is given by \(E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)
Here, \(r=3.0 \mathrm{~m}\)
\(
\therefore \quad E=\left(9.0 \times 10^9\right) \times \frac{2 \times 10^{-7}}{(3.0)^2}=200 \mathrm{~N} / \mathrm{C}
\)

Example 40: A small conducting spherical shell with inner radius \(a\) and outer radius \(b\) is concentric with a larger conducting spherical shell with inner radius \(c\) and outer radius \(d\). The inner shell has a total charge \(+2 q\) and the outer shell has a charge \(+4 q\).
(i) What is the total charge on the (a) inner surface of the small shell, (b) outer surface of the small shell, (c) inner surface of the large shell (d) and outer surface of the large shell?
(ii) Calculate the electric field in terms of \(q\) and the distance \(r\) from the common centre of two shells for (a) \(r<a\), (b) \(a<r<b\), (c) \(b<r<c\), (d) \(c<r<d\) (e) and \(r>d\).

Solution: Charge distribution is shown below.

(i) Total charge on inner shell \(=2 q\)
Total charge on outer shell \(=4 q\)
Charge distribution
(a) Charge on inner surface of small shell \(=0\) (inner surface has no charge)
(b) Charge on outer surface of small shell \(=2 q\)
(c) Charge on inner surface of large shell \(=-2 q\) (facing surface have equal and opposite charges)
(d) Charge on outer surface of large shell \(=6 q\) (total charge on outer shell is \(4 q\) )
(ii) To calculate electric field, draw a sphere with centre \(O\) through that point, where electric field is required.
Assume charge to be concentrated at centre and apply formula of point charge.
(a) \(r<a\), enclosed charge \(=0, E=0\)
(b) \(a<r<b\), enclosed charge \(=0, E=0\) or electric field inside conductor \(=0\)
(c) \(b<r<c\), enclosed charge \(=2 q, E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 q}{r^2}\)
(d) \(c<r<d\), enclosed charge \(=0, E=0\), or electric field inside conductor \(=0\)
(e) \(r>d\), enclosed charge \(=6 q, E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{6 q}{r^2}\)

Electric field due to a non-conducting solid sphere of charge

Suppose positive charge \(q\) is uniformly distributed throughout the volume of a non-conducting solid sphere of radius \(R\).

At an internal point

For finding the electric field at a distance \(r(<R)\) from the centre, let us choose our Gaussian surface a sphere of radius \(r\), concentric with the charge distribution. From symmetry, the magnitude of an electric field \(E\) has the same value at every point on the Gaussian surface and the direction of \(\mathbf{E}\) is radial at every point on the surface. So, applying Gauss’s law, we have
\(
E S=\frac{q_{\mathrm{enc}}}{\varepsilon_0}
\)
Here, \(\quad S=4 \pi r^2\) and \(q_{\text {enc }}=(\rho)\left(\frac{4}{3} \pi r^3\right)\)
Here, \(\rho=\) charge per unit volume \(=\frac{q}{(4 / 3) \pi R^3}\)
Substituting these values in Eq. (i), we get
\(
E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R^3} \cdot r
\)
\(
E \propto r
\)
At the centre of sphere, \(r=0, \quad\) so \(\quad E=0\)
At the surface of sphere, \(r=R, \quad\) so \(\quad E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R^2}\)

At an external point

To find the electric field outside the charged sphere, we use a spherical Gaussian surface of radius \(r(>R)\). This surface encloses the entire charged sphere, so \(q_{\text {in }}=q\) and Gauss’s law gives,
\(
\begin{aligned}
E\left(4 \pi r^2\right) & =\frac{q}{\varepsilon_0} \\
E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} \text { or } E \propto \frac{1}{r^2}
\end{aligned}
\)
Thus, for a uniformly charged solid sphere, we have the following formulae for the magnitude of the electric field.
\(
\begin{aligned}
& E_{\text {inside }}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R^3} \cdot r \\
& E_{\text {surface }}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R^2} \\
& E_{\text {outside }}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}
\end{aligned}
\)
Thus, the electric field at any external point is the same as, if the total charge is concentrated at centre.
The variation of an electric field \((E)\) with the distance from the centre of the sphere \((r)\) is shown in the figure.

Note: If we set \(r=R\) in either of the two expressions for \(E\) (outside and inside the sphere), we get the same result, \(E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R^2}\)
this is because \(E\) is continuous function of \(r\) in this case. By contrast, for the charged conducting sphere, the magnitude of the electric field is discontinuous at \(r=R\) (it jumps from \(E=0\) to \(\left.E=\sigma / \varepsilon_0\right)\).

Example 41: At a point 20 cm from the centre of a uniformly charged dielectric sphere of radius 10 cm, the electric field is \(100 \mathrm{~V} / \mathrm{m}\). Find the electric field at 3 cm from the centre of the sphere.

Solution:  Electric field outside the dielectric sphere, \(E_{\text {out }}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2}\)
The electric field inside the dielectric sphere, \(E_{\text {in }}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R^3} x\)
\(
\begin{array}{ll}
\therefore & E_{\text {in }}=E_{\text {out }} \times \frac{r^2 x}{R^3} \\
\Rightarrow & E=100 \times \frac{3 \times(20)^2}{10^3}=120 \mathrm{~V} / \mathrm{m}
\end{array}
\)

Example 42: Two non-conducting spheres of radius \(R\) have charge \(Q\) uniformly distributed on them. The centres of spheres are at \(x=0\) and \(x=3 R\). Find the magnitude and direction of the net electric field on the \(X\)-axis at
(i) \(x=0\),
(ii) \(x=\frac{R}{2}\),
(iii) \(x=\frac{3 R}{2}\)
(iv) and \(x=4 R\).

Solution: Electric field inside sphere at distance \(r\) from centre
\(
=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q r}{R^3}
\)

On the surface or outside, the whole charge is assumed to be concentrated at centre.
(i) At \(x=0, E_1=0, E_2=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{(3 R)^2}\), towards left \(E_{\text {net }}=E_2=\frac{Q}{36 \pi \varepsilon_0 R^2}\), along \(-X\)-axis
(ii) At \(\begin{aligned} x & =R / 2, \\ E_1 & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q R / 2}{R^3}=\frac{Q}{8 \pi \varepsilon_0 R^2}, \text { along }+X \text {-axis } \\ E_2 & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{(5 R / 2)^2}=\frac{Q}{25 \pi \varepsilon_0 R^2}, \text { along }-X \text {-axis } \\ E_{\text {net }} & =E_1-E_2=\frac{17 Q}{200 \pi \varepsilon_0 R^2}, \text { along }+X \text {-axis }\end{aligned}\)
(iii) At \(x=3 R / 2, E_{\text {net }}=0 \left(\because E_1=E_2\right)\)
(iv) At \(\begin{aligned} x & =4 R, E_1=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{(4 R)^2}, E_2=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2} \\ E_{\text {net }} & =E_1+E_2=\frac{17 Q}{64 \pi \varepsilon_0 R^2}, \text { along }+X \text {-axis }\end{aligned}\)

Example 43: A non-conducting sphere of radius \(R\) has a spherical cavity of radius \(R / 2\) as shown in the figure. The solid part of the sphere has a uniform volume charge density \(\rho\). Find the magnitude and direction of the electric field at point (a) \(O\) and (b) A.

Solution: For a non-conducting sphere of radius \(R\) having volume charge density \(\rho\).

(i) At \(P, r<R\) (inside), \(E=\frac{\rho r}{3 \varepsilon_0}\)
(ii) At \(S, r>R\) (outside), \(E=\frac{\rho R^3}{3 \varepsilon_0 r^2}\)
(iii) At \(Q, r=R\) (surface), \(E=\frac{\rho R}{3 \varepsilon_0}\)
The given sphere can be shown as,

(a) At \(O, E_1=0, E_2=\frac{\rho R / 2}{3 \varepsilon_0}=\frac{\rho R}{6 \varepsilon_0}\), towards left \(\Rightarrow \quad E_0=\frac{\rho R}{6 \varepsilon_0}\), towards left
(b) At \(A, E_1=\frac{\rho R}{3 \varepsilon_0}\), towards right \(E_2=\frac{\rho(R / 2)^3}{3 \varepsilon_0(3 R / 2)^2}=\frac{\rho R}{54 \varepsilon_0}\), towards left \(\therefore \quad E_A=E_1-E_2=\frac{\rho R \mid}{\varepsilon_0}\left[\frac{1}{3}-\frac{1}{54}\right]=\frac{17 \rho R}{54 \varepsilon_0}\), towards right

Example 44: At a far away distance \(r\) along the axis from an electric dipole electric field is \(\mathbf{E}\). Find the electric field at distance \(2 r\) along the perpendicular bisector.

Solution: Along the axis of the dipole,
\(
E=\frac{1}{4 \pi \varepsilon_0} \frac{2 p}{r^3} \dots(i)
\)
This electric field is in the direction of \(\mathbf{p}\) Along the perpendicular bisector at a distance \(2 r\),
\(
E^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{p}{(2 r)^3} \dots(ii)
\)
From Eqs. (i) and (ii), we can see that
\(
E^{\prime}=\frac{E}{16}
\)
Moreover, \(E^{\prime}\) is in the opposite direction of \(\mathbf{p}\) Hence,
\(
\mathbf{E}^{\prime}=-\frac{\mathbf{E}}{16}
\)