NCERT Exercise Q & A

NCERT EXERCISE PROBLEMS

Q1.1: What is the force between two small charged spheres having charges of \(2 \times 10^{-7} \mathrm{C}\) and \(3 \times 10^{-7} \mathrm{C}\) placed 30 cm apart in air?

Answer: Charge on the first sphere, \(q_1=2 \times 10^{-7} \mathrm{C}\)
Charge on the second sphere, \(q_2=3 \times 10^{-7} \mathrm{C}\)
Distance between the spheres, \(r=30 \mathrm{~cm}=0.3 \mathrm{~m}\)
Electrostatic force between the spheres is given by the relation,
\(
F=\frac{q_1 q_2}{4 \pi \epsilon_0 r^2}
\)
Where, \(\epsilon_0=\) Permittivity of free space
\(
\begin{aligned}
& \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2} \\
& F=\frac{9 \times 10^9 \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^2}=6 \times 10^{-3} \mathrm{~N}
\end{aligned}
\)
Hence, force between the two small charged spheres is \(6 \times 10^{-3} \mathrm{~N}\). The charges are of same nature. Hence, force between them will be repulsive.

Q1.2: The electrostatic force on a small sphere of charge \(0.4 \mu \mathrm{C}\) due to another small sphere of charge \(-0.8 \mu \mathrm{C}\) in air is 0.2 N . (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Answer: (a) Electrostatic force on the first sphere, \(F=0.2 \mathrm{~N}\)
Charge on this sphere, \(q_1=0.4 \mu \mathrm{C}=0.4 \times 10^{-6} \mathrm{C}\)
Charge on the second sphere, \(q_2=-0.8 \mu \mathrm{C}=-0.8 \times 10^{-6} \mathrm{C}\)
Electrostatic force between the spheres is given by the relation,
\(
F=\frac{q_1 q_2}{4 \pi \epsilon_0 r^2}
\)
Where, \(\epsilon_0=\) Permittivity of free space
\(
\begin{aligned}
& \text { And, } \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2} \\
& \begin{aligned}
r^2 & =\frac{q_1 q_2}{4 \pi \epsilon_0 F} \\
& =\frac{0.4 \times 10^{-6} \times 8 \times 10^{-6} \times 9 \times 10^9}{0.2} \\
& =144 \times 10^{-4} \\
r & =\sqrt{144 \times 10^{-4}}=0.12 \mathrm{~m}
\end{aligned}
\end{aligned}
\)
The distance between the two spheres is 0.12 m .
(b) Using Newton’s third law, the force exerted by the spheres on each other will be equal in magnitude. Therefore, the force on the second sphere due to the first is 0.2 N (This force will be attractive since charges are of opposite sign).

Q1.3: Check that the ratio \(k e^2 / G m_e m_p\) is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Answer: The given ratio is \(\frac{k e^2}{\mathrm{G} m_{\mathrm{e}} m_{\mathrm{p}}}\).
Where,
\(\mathrm{G}=\) Gravitational constant
Its unit is \(N \mathrm{~m}^2 \mathrm{~kg}^{-2}\).
\(m_{\mathrm{e}}\) and \(m_{\mathrm{p}}=\) Masses of electron and proton.
Their unit is kg .
\(e=\) Electric charge.
Its unit is C.
\(k=\mathrm{A}\) constant
\(
=\frac{1}{4 \pi \epsilon_0}
\)
\(\epsilon_0=\) Permittivity of free space
Its unit is \(\mathrm{N} \mathrm{m}^2 \mathrm{C}^{-2}\).
Therefore, unit of the given ratio \(\begin{aligned} \frac{k e^2}{\mathrm{G} m_{\mathrm{e}} m_{\mathrm{p}}} & =\frac{\left[\mathrm{Nm}^2 \mathrm{C}^{-2}\right]\left[\mathrm{C}^{-2}\right]}{\left[\mathrm{Nm}^2 \mathrm{~kg}^{-2}\right][\mathrm{kg}][\mathrm{kg}]} \\ & =\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\end{aligned}\)
Hence, the given ratio is dimensionless.
\(
\begin{aligned}
& \mathrm{e}=1.6 \times 10^{-19} \mathrm{C} \\
& \mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^2 \mathrm{~kg}^{-2} \\
& m_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg} \\
& m_{\mathrm{p}}=1.66 \times 10^{-27} \mathrm{~kg}
\end{aligned}
\)
Hence, the numerical value of the given ratio is
\(
\frac{k e^2}{\mathrm{G} m_e m_p}=\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{6.67 \times 10^{-11} \times 9.1 \times 10^{-3} \times 1.67 \times 10^{-22}} \approx 2.3 \times 10^{39}
\)
This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.

Q1.4: (a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Answer: (a) Electric charge of a body is quantized. This means that only integral \((1,2, \ldots, n)\) number of electrons can be transferred from one body to the other. Charges are not transferred in fraction. Hence, a body possesses total charge only in integral multiples of electric charge.
(b) In macroscopic or large scale charges, the charges used are huge as compared to the magnitude of electric charge. Hence, quantization of electric charge is of no use on macroscopic scale. Therefore, it is ignored and it is considered that electric charge is continuous.

Q1.5: When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Answer: Rubbing produces charges of equal magnitude but of opposite nature on the two bodies because charges are created in pairs. This phenomenon of charging is called charging by friction. The net charge on the system of two rubbed bodies is zero. This is because equal amount of opposite charges annihilate each other. When a glass rod is rubbed with a silk cloth, opposite natured charges appear on both the bodies. This phenomenon is in consistence with the law of conservation of energy. A similar phenomenon is observed with many other pairs of bodies.

Q1.6: Four point charges \(q_{\mathrm{A}}=2 \mu \mathrm{C}, q_{\mathrm{B}}=-5 \mu \mathrm{C}, q_{\mathrm{C}}=2 \mu \mathrm{C}\), and \(q_{\mathrm{D}}=-5 \mu \mathrm{C}\) are located at the corners of a square \(A B C D\) of side 10 cm . What is the force on a charge of \(1 \mu \mathrm{C}\) placed at the centre of the square?

Answer: The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.

Where,
(Sides) \(A B=B C=C D=A D=10 \mathrm{~cm}\)
(Diagonals) \(A C=B D=10 \sqrt{2} \mathrm{~cm}\)
\(
\mathrm{AO}=\mathrm{OC}=\mathrm{DO}=\mathrm{OB}=5 \sqrt{2} \mathrm{~cm}
\)
\(\mathrm{A}\) charge of amount \(1 \mu \mathrm{C}\) is placed at point \(\mathrm{O}\).
Force of repulsion between charges placed at corner \(A\) and centre \(\mathrm{O}\) is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner \(\mathrm{C}\) and centre \(\mathrm{O}\). Hence, they will cancel each other.
Similarly, force of attraction between charges placed at corner \(\mathrm{B}\) and centre \(\mathrm{O}\) is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner \(\mathrm{D}\) and centre \(\mathrm{O}\). Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on \(1 \mu \mathrm{C}\) charge at centre \(\mathrm{O}\) is zero.

Q1.7: (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
(b) Explain why two field lines never cross each other at any point?

Answer: (a) An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.
(b) If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.

Q1.8: Two point charges \(q_{\mathrm{A}}=3 \mu \mathrm{C}\) and \(q_{\mathrm{B}}=-3 \mu \mathrm{C}\) are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint \(O\) of the line \(A B\) joining the two charges?
(b) If a negative test charge of magnitude \(1.5 \times 10^{-9} \mathrm{C}\) is placed at this point, what is the force experienced by the test charge?

Answer: (a) The situation is represented in the given figure. \(O\) is the mid-point of line \(A B\).

Distance between the two charges, \(A B=20 \mathrm{~cm}\)
\(
\therefore A O=O B=10 \mathrm{~cm}
\)
Net electric field at point \(\mathrm{O}=E\)
Electric field at point \(O\) caused by \(+3 \mu \mathrm{C}\) charge,
\(
E_1=\frac{3 \times 10^{-6}}{4 \pi \epsilon_0(\mathrm{AO})^2}=\frac{3 \times 10^{-6}}{4 \pi \epsilon_0\left(10 \times 10^{-2}\right)^2} \mathrm{~N} / \mathrm{C} \quad \text { along } O B
\)
Where,
\(
\begin{aligned}
& \epsilon_0=\text { Permittivity of free space } \\
& \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}
\end{aligned}
\)
Magnitude of electric field at point O caused by \(-3 \mu \mathrm{C}\) charge,
\(
E_2=\left|\frac{-3 \times 10^{-6}}{4 \pi \epsilon_0(\mathrm{OB})^2}\right|_{=} \frac{3 \times 10^{-6}}{4 \pi \epsilon_0\left(10 \times 10^{-2}\right)^2} \mathrm{~N} / \mathrm{C} \quad \text { along } O B
\)
\(
\therefore E=E_1+E_2
\)
\(
=2 \times\left[\left(9 \times 10^9\right) \times \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^2}\right] \text { [Since the values of [latex]E_1[latex] and [latex]E_2[latex] are same, the value is multiplied with 2] }
\)
\(
\begin{aligned}
&=5.4 \times 10^6 \mathrm{~N} / \mathrm{C} \text { along } \mathrm{OB}\\
&\text { Therefore, the electric field at mid-point } \mathrm{O} \text { is } 5.4 \times 10^6 \mathrm{~N} \mathrm{C}^{-1} \text { along } \mathrm{OB} \text {. }
\end{aligned}
\)

(b) A test charge of amount \(1.5 \times 10^{-9} \mathrm{C}\) is placed at mid-point O .
\(
q=1.5 \times 10^{-9} \mathrm{C}
\)
Force experienced by the test charge \(=F\)
\(
\begin{aligned}
& \therefore F=q E \\
& =1.5 \times 10^{-9} \times 5.4 \times 10^6 \\
& =8.1 \times 10^{-3} \mathrm{~N}
\end{aligned}
\)
The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point \(B\) but attracted towards point \(A\).
Therefore, the force experienced by the test charge is \(8.1 \times 10^{-3} \mathrm{~N}\) along OA.

Q1.9: A system has two charges \(q_{\mathrm{A}}=2.5 \times 10^{-7} \mathrm{C}\) and \(q_{\mathrm{B}}=-2.5 \times 10^{-7} \mathrm{C}\) located at points \(A:(0,0,-15 \mathrm{~cm})\) and \(B:(0,0,+15 \mathrm{~cm})\), respectively. What are the total charge and electric dipole moment of the system?

Answer: Both the charges can be located in a coordinate frame of reference as shown in the given figure.

At A, amount of charge, \(q_{\mathrm{A}}=2.5 \times 10^{-7} \mathrm{C}\)
At B , amount of charge, \(q_{\mathrm{B}}=-2.5 \times 10^{-7} \mathrm{C}\)
Total charge of the system,
\(
\begin{aligned}
& q=q_{\mathrm{A}}+q_{\mathrm{B}} \\
& =2.5 \times 10^7 \mathrm{C}-2.5 \times 10^{-7} \mathrm{C} \\
& =0
\end{aligned}
\)
Distance between two charges at points A and B,
\(
d=15+15=30 \mathrm{~cm}=0.3 \mathrm{~m}
\)
Electric dipole moment of the system is given by,
\(
\begin{aligned}
& p=q_{\mathrm{A}} \times d=q_{\mathrm{B}} \times d \\
& =2.5 \times 10^{-7} \times 0.3 \\
& =7.5 \times 10^{-3} \mathrm{C} \text { m along positive } z \text {-axis }
\end{aligned}
\)
Therefore, the electric dipole moment of the system is \(7.5 \times 10^{-8} \mathrm{C} \mathrm{m}\) along positive \(z\)-axis.

Q1.10: An electric dipole with dipole moment \(4 \times 10^{-9} \mathrm{C} \mathrm{m}\) 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.

Answer: Electric dipole moment, \(p=4 \times 10^{-9} \mathrm{C} \mathrm{m}\)
Angle made by \(p\) with a uniform electric field, \(\theta=30^{\circ}\)
Electric field, \(E=5 \times 10^4 \mathrm{~N} \mathrm{C}^{-1}\)
Torque acting on the dipole is given by the relation,
\(
\begin{aligned}
& T=p E \sin \theta \\
& =4 \times 10^{-9} \times 5 \times 10^4 \times \sin 30 \\
& =20 \times 10^{-5} \times \frac{1}{2} \\
& =10^{-4} \mathrm{Nm}
\end{aligned}
\)
Therefore, the magnitude of the torque acting on the dipole is \(10^{-4} \mathrm{~N} \mathrm{~m}\).

Q1.11: A polythene piece rubbed with wool is found to have a negative charge of \(3 \times 10^{-7} \mathrm{C}\).
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?

Answer:(a) When polythene is rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene becomes negatively charged.
Amount of charge on the polythene piece, \(q=-3 \times 10^{-7} \mathrm{C}\)
Amount of charge on an electron, \(e=-1.6 \times 10^{-19} \mathrm{C}\)
Number of electrons transferred from wool to polythene \(=n\)
\(n\) can be calculated using the relation,
\(
\begin{aligned}
& q=n e \\
& n=\frac{q}{e} \\
& =\frac{-3 \times 10^{-7}}{-1.6 \times 10^{-19}} \\
& =1.87 \times 10^{12}
\end{aligned}
\)
Therefore, the number of electrons transferred from wool to polythene is \(1.87 \times\) \(10^{12}\).

(b) Yes.
There is a transfer of mass taking place. This is because an electron has mass,
\(
m_e=9.1 \times 10^{-3} \mathrm{~kg}
\)
Total mass transferred to polythene from wool,
\(
\begin{aligned}
& m=m_e \times n \\
& =9.1 \times 10^{-31} \times 1.85 \times 10^{12} \\
& =1.706 \times 10^{-18} \mathrm{~kg}
\end{aligned}
\)
Hence, a negligible amount of mass is transferred from wool to polythene.

Q1.12: (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm . What is the mutual force of electrostatic repulsion if the charge on each is \(6.5 \times 10^{-7} \mathrm{C}\) ? The radii of \(A\) and \(B\) are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Answer: (a) Charge on sphere \(\mathrm{A}, q_{\mathrm{A}}=\) Charge on sphere \(\mathrm{B}, q_{\mathrm{B}}=6.5 \times 10^{-7} \mathrm{C}\)
Distance between the spheres, \(r=50 \mathrm{~cm}=0.5 \mathrm{~m}\)
Force of repulsion between the two spheres,
\(
F=\frac{q_{\mathrm{A}} q_{\mathrm{B}}}{4 \pi \epsilon_0 r^2}
\)
Where,
\(\epsilon_0=\) Free space permittivity
\(
\begin{aligned}
& \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2} \\
& F=\frac{9 \times 10^9 \times\left(6.5 \times 10^{-7}\right)^2}{(0.5)^2} \\
& \therefore \\
& =1.52 \times 10^{-2} \mathrm{~N}
\end{aligned}
\)
Therefore, the force between the two spheres is \(1.52 \times 10^{-2} \mathrm{~N}\).

(b) After doubling the charge, charge on sphere \(\mathrm{A}, q_{\mathrm{A}}=\) Charge on sphere \(\mathrm{B}, q_{\mathrm{B}}=2\)
\(
\times 6.5 \times 10^{-7} \mathrm{C}=1.3 \times 10^{-6} \mathrm{C}
\)
The distance between the spheres is halved.
\(
\therefore r=\frac{0.5}{2}=0.25 \mathrm{~m}
\)
Force of repulsion between the two spheres,
\(
\begin{aligned}
& F=\frac{q_{\mathrm{A}} q_{\mathrm{B}}}{4 \pi \epsilon_0 r^2} \\
& =\frac{9 \times 10^9 \times 1.3 \times 10^{-6} \times 1.3 \times 10^{-6}}{(0.25)^2} \\
& =16 \times 1.52 \times 10^{-2} \\
& =0.243 \mathrm{~N}
\end{aligned}
\)
Therefore, the force between the two spheres is 0.243 N.

Q1.13: Figure 1.30 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Answer: Opposite charges attract each other and same charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.
The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.

Q1.14: Consider a uniform electric field \(\mathbf{E}=3 \times 10^3 \hat{\mathbf{i}} \mathrm{~N} / \mathrm{C}\). (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the \(y z\) plane? (b) What is the flux through the same square if the normal to its plane makes a \(60^{\circ}\) angle with the \(x\)-axis?

Answer: (a) Electric field intensity, \(\vec{E}=3 \times 10^3 i \mathrm{~N} / \mathrm{C}\)
Magnitude of electric field intensity, \(|\vec{E}|=3 \times 10^3 \mathrm{~N} / \mathrm{C}\)
Side of the square, \(s=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Area of the square, \(A=s^2=0.01 \mathrm{~m}^2\)
The plane of the square is parallel to the \(y-z\) plane [therefore its normal is in x -direction.(i.e \(\widehat{i}\) direction)]. Hence, the angle between the unit vector normal to the plane and electric field, \(\theta=0^{\circ}\)
Flux ( \(\Phi\) ) through the plane is given by the relation,
\(
\begin{aligned}
& \Phi=|\vec{E}| A \cos \theta \\
& =3 \times 10^3 \times 0.01 \times \cos 0^{\circ} \\
& =30 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}
\end{aligned}
\)

(b) Now, Since the normal of the square plane makes an angle of \(60^{\circ}\) with the \(x\)-axis. Hence, \(\theta=60^{\circ}\)
\(
\begin{aligned}
& \text { Flux, } \Phi=|\vec{E}| A \cos \theta \\
& =3 \times 10^3 \times 0.01 \times \cos 60^{\circ} \\
& =30 \times \frac{1}{2}=15 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}
\end{aligned}
\)

Q1.15: What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Answer: All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.

Alternatively, using Gauss’s law, we know that the flux of electric field through any closed surface S is \(1 / \epsilon_0\) times the total charge enclosed by S.
i.e. \(\phi=q / \epsilon_0\)
where, \(\mathrm{q}=\) net charge enclosed and \(\epsilon_0=\) permittivity of free space (constant)
Since there is no charge enclosed in the cube, hence \(\phi=0\).

Q1.16: Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is \(8.0 \times 10^3 \mathrm{Nm}^2 / \mathrm{C}\). (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Answer: (a) Net outward flux through the surface of the box, \(\Phi=8.0 \times 10^3 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}\)
For a body containing net charge \(a\), flux is given by the relation (Using Gauss’s law,),
\(
\begin{aligned}
& \phi=\frac{q}{\epsilon_0} \\
& \epsilon_0=\text { Permittivity of free space } \\
& =8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^2 \mathrm{~m}^{-2} \\
& q=\epsilon_0 \Phi \\
& =8.854 \times 10^{-12} \times 8.0 \times 10^3 \\
& =7.08 \times 10^{-8} \\
& =0.07 \mu \mathrm{C}
\end{aligned}
\)
Therefore, the net charge inside the box is \(0.07 \mu \mathrm{C}\).
(b) No, Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.

Alternatively, Using Gauss’s law, we know that \(\phi=q / \epsilon_0\)
Since flux is zero, \(q=0\), but this \(q\) is the net charge enclosed by the surface.
Hence, we can conclusively say that net charge is zero, but we cannot conclude that there are no charges inside the box.

Q1.17: A point charge \(+10 \mu \mathrm{C}\) is a distance 5 cm directly above the centre of a square of side 10 cm , as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm .)

Answer: The square can be considered as one face of a cube of edge 10 cm with a centre where charge \(q\) is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces.
\(
\begin{aligned}
&\phi_{\text {Total }}=\frac{q}{\epsilon_0}\\
&\text { Hence, electric flux through one face of the cube i.e., through the square, } \quad \phi=\frac{\phi_{\text {Toatal }}}{6}\\
&=\frac{1}{6} \frac{q}{\epsilon_0}
\end{aligned}
\)
Where,
\(
\begin{aligned}
& \in_0=\text { Permittivity of free space } \\
& =8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^2 \mathrm{~m}^{-2} \\
& q=10 \mu \mathrm{C}=10 \times 10^{-6} \mathrm{C} \\
& \quad \phi=\frac{1}{6} \times \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}} \\
& =1.88 \times 10^5 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}
\end{aligned}
\)
Therefore, electric flux through the square is \(1.88 \times 10^5 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}\).

Q1.18: A point charge of \(2.0 \mu \mathrm{C}\) is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Answer: Net electric flux ( \(\phi_{\mathrm{Net}}\) ) through the cubic surface is given by,
\(
\phi_{\mathrm{Net}}=\frac{q}{\epsilon_0}
\)
Where,
\(\epsilon_0=\) Permittivity of free space
\(
=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^2 \mathrm{~m}^{-2}
\)
\(q=\) Net charge contained inside the cube \(=2.0 \mu \mathrm{C}=2 \times 10^{-6} \mathrm{C}\)
\(
\begin{aligned}
& \quad \phi_{\text {Net }}=\frac{2 \times 10^{-6}}{8.854 \times 10^{-12}} \\
& =2.26 \times 10^5 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}
\end{aligned}
\)
The net electric flux through the surface is \(2.26 \times 10^5 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}\).

Q1.19: A point charge causes an electric flux of \(-1.0 \times 10^3 \mathrm{Nm}^2 / \mathrm{C}\) to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

Answer: (a) Electric flux, \(\Phi=-1.0 \times 10^3 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}\)
Radius of the Gaussian surface,
\(
r=10.0 \mathrm{~cm}
\)
Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., \(-10^3 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}\).
(b) Electric flux is given by the relation,
\(
\phi=\frac{q}{\epsilon_0}
\)
Where,
\(q=\) Net charge enclosed by the spherical surface
\(\epsilon_0=\) Permittivity of free space \(=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^2 \mathrm{~m}^{-2}\)
\(
\begin{aligned}
& \therefore q=\phi \in_0 \\
& =-1.0 \times 10^3 \times 8.854 \times 10^{-12} \\
& =-8.854 \times 10^{-9} \mathrm{C} \\
& =-8.854 \mathrm{~nC}
\end{aligned}
\)
Therefore, the value of the point charge is \(-8.854 \mathrm{~nC}\).

Q1.20: A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is \(1.5 \times 10^3 \mathrm{~N} / \mathrm{C}\) and points radially inward, what is the net charge on the sphere?

Answer: Electric field intensity \((E)\) at a distance \((d)\) from the centre of a sphere containing net charge \(q\) is given by the relation,
\(
E=\frac{q}{4 \pi \epsilon_0 d^2}
\)
Where,
\(q=\) Net charge \(=1.5 \times 10^3 \mathrm{~N} / \mathrm{C}\)
\(d=\) Distance from the centre \(=20 \mathrm{~cm}=0.2 \mathrm{~m}\)
\(\epsilon_0=\) Permittivity of free space
And, \(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}\)
\(
q=E\left(4 \pi \epsilon_0\right) d^2
\)
\(
\begin{aligned}
&\begin{aligned}
& =\frac{1.5 \times 10^3 \times(0.2)^2}{9 \times 10^9} \\
& =6.67 \times 10^9 \mathrm{C} \\
& =6.67 \mathrm{nC}
\end{aligned}\\
&\text { Therefore, charge on the conducting sphere is }-6.67 \mathrm{nC} \text { (since flux is inwards) }
\end{aligned}
\)

Q1.21: A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of \(80.0 \mu \mathrm{C} / \mathrm{m}^2\). (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Answer: 

(a) Diameter of the sphere, \(d=2.4 \mathrm{~m}\)
Radius of the sphere, \(r=1.2 \mathrm{~m}\)
Surface charge density, \(\sigma=80.0 \mu \mathrm{C} / \mathrm{m}^2=80 \times 10^{-6} \mathrm{C} / \mathrm{m}^2\)
Total charge on the surface of the sphere,
\(
\begin{aligned}
& Q=\text { Charge density } \times \text { Surface area } \\
& =\sigma \times 4 \pi r^2 \\
& =80 \times 10^{-6} \times 4 \times 3.14 \times(1.2)^2 \\
& =1.447 \times 10^{-3} \mathrm{C}
\end{aligned}
\)
Therefore, the charge on the sphere is \(1.447 \times 10^{-3} \mathrm{C}\).

(b) Total electric flux ( \(\phi_{\text {Total }}\) ) leaving out the surface of a sphere containing net charge \(Q\) is given by the relation,
\(
\phi_{\text {Total }}=\frac{Q}{\epsilon_0}
\)
Where,
\(
\begin{aligned}
& \epsilon_0=\text { Permittivity of free space } \\
& =8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^2 \mathrm{~m}^{-2} \\
& Q=1.447 \times 10^{-3} \mathrm{C} \\
& \phi_{\text {Total }}=\frac{1.44 \times 10^{-3}}{8.854 \times 10^{-12}} \\
& =1.63 \times 10^3 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2
\end{aligned}
\)
Therefore, the total electric flux leaving the surface of the sphere is \(1.63 \times 10^3 \mathrm{~N} \mathrm{C}^{-1}\) \(m^2\)

Q1.22: An infinite line charge produces a field of \(9 \times 10^4 \mathrm{~N} / \mathrm{C}\) at a distance of 2 cm . Calculate the linear charge density.

Answer: Electric field produced by the infinite line charges at a distance \(d\) having linear charge density \(\lambda\) is given by the relation,
\(
\begin{aligned}
& E=\frac{\lambda}{2 \pi \epsilon_0 d} \\
& \lambda=2 \pi \epsilon_0 d E
\end{aligned}
\)
Where,
\(
\begin{aligned}
& d=2 \mathrm{~cm}=0.02 \mathrm{~m} \\
& E=9 \times 10^4 \mathrm{~N} / \mathrm{C} \\
& \epsilon_0=\text { Permittivity of free space } \\
& \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2} \\
& \lambda=\frac{0.02 \times 9 \times 10^4}{2 \times 9 \times 10^9} \\
& =10 \mu \mathrm{C} / \mathrm{m}
\end{aligned}
\)
Therefore, the linear charge density is \(10 \mu \mathrm{C} / \mathrm{m}\).

Q1.23: Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude \(17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^2\). What is \(\mathbf{E}\) : (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

Answer:The situation is represented in the following figure.

\(A\) and \(B\) are two parallel plates close to each other. Outer region of plate \(A\) is labelled as \(\mathbf{I}\), outer region of plate \(B\) is labelled as \(\mathbf{III}\), and the region between the plates, \(A\) and \(B\), is labelled as \(\mathbf{II}\).
Charge density of plate \(\mathrm{A}, \sigma=17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^2\)
Charge density of plate \(B, \sigma=-17.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^2\)
In the regions, I and III, electric field \(E\) is zero. This is because charge is not enclosed by the respective plates.
Electric field \(E\) in region II is given by the relation,
\(
E=\frac{\sigma}{\epsilon_0}
\)
Where,
\(
\begin{aligned}
&\begin{aligned}
& \epsilon_0=\text { Permittivity of free space }=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^2 \mathrm{~m}^{-2} \\
& E=\frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}} \\
& =1.92 \times 10^{-10} \mathrm{~N} / \mathrm{C}
\end{aligned}\\
&\text { Therefore, electric field between the plates is } 1.92 \times 10^{-10} \mathrm{~N} / \mathrm{C} \text {. }
\end{aligned}
\)

NCERT EXEMPLAR PROBLEMS

Very Short Questions

Q1.1: An arbitrary surface encloses a dipole. What is the electric flux through this surface?

Answer: According to Gauss’ law, the electric flux through an enclosed surface is given by \(\oint_s \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{S}}=\frac{q_{\text {enclosed }}}{\mathcal{E}_0}\).
The net charge on a dipole is given by \(-q+q=0\), hence \(q_{\text {enclosed }}=0\)
Hence the electric flux through a surface enclosing a dipole \(=\frac{-q+q}{\varepsilon_0}=\frac{q_{\text {enclosed }}}{\varepsilon_0}=0\)

Q1.2: A metallic spherical shell has an inner radius \(R_1\) and outer radius \(R_2\). A charge \(Q\) is placed at the centre of the spherical cavity. What will be surface charge density on (i) the inner surface, and (ii) the outer surface?

Answer: A charge \(Q\) is placed at the centre of the spherical cavity. So, the charge induced at the inner surface of the sphere will be \(-Q\) and at outer surface of the sphere is \(+Q\).

The surface charge density on the inner surface \(\sigma_1=\frac{\text { Charge }}{\text { Area }}=\frac{-Q}{4 \pi R_1^2}\)
Surface charge density on the outer surface \(\sigma_2=\frac{+Q}{4 \pi R_2^2}\)
Hence,
(i) \(=\frac{-Q}{4 \pi R_1^2}\)
(ii) \(\frac{+Q}{4 \pi R_2^2}\)

Q1.3: The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero?

Answer: In any neutral atom, the number of electrons and protons are equal, and the protons and electrons are bound into an atom with distinct and independent existence. Electrostatic fields are caused by the presence of excess charges. But there can be no excess charge on the intersurface of an isolated conductor. So, the electrostatic fields inside a conductor is zero despite the fact that the dimensions of an atom are of the order of an Angstrom.

Q1.4: If the total charge enclosed by a surface is zero, does it imply that the elecric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero.

Answer: According to Gauss’ law, the flux associated with any closed surface is given by \(\oint_s \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{S}}=\frac{q_{\text {enclosed }}}{\varepsilon_0}.\). The term \(q_{\text {enclosed }}\) on the right side of the equation includes the sum of all charges enclosed by the surface called (Gaussian surface). In left side equation,the electric field is due to all the charges present both inside as well as outside the Gaussian surface. Thus, despite being total charge enclosed by a surface zero, it doesn’t imply that the electric field everywhere on the surface is zero, the field may be normal to the surface.  Also, conversely if the electric field everywhere on a surface is zero, it doesn’t imply that net charge inside it is zero.
i.e., \(\text { Putting } E=0 \text { in } \oint \text { E.dS }=\frac{q}{\varepsilon_0}\)
we get \(q=0\)

Q1.5: Sketch the electric field lines for a uniformly charged hollow cylinder shown in Fig 1.8.

Answer: The electric field lines starts from positive charges and move towards infinity and meet plane surface normally as shown in the figure below:

Important point: No electric field lines will be present inside the cylinder because of electrostatic shielding. Electrostatic shielding/screening is the phenomenon of protecting a certain region of space from external electric field. Sensitive instruments and appliances are affected seriously with strong external electrostatic fields. Their working suffers and they may start misbehaving under the effect of unwanted fields. The electrostatic shielding can be achieved by protecting and enclosing the sensitive instruments inside hollow conductor because inside hollow conductors, electric fields is zero.

Q1.6: What will be the total flux through the faces of the cube (Fig. 1.9) with side of length \(a\) if a charge \(q\) is placed at
(a) A : a corner of the cube.
(b) B : mid-point of an edge of the cube.
(c) C : centre of a face of the cube.
(d) D: mid-point of B and C.

Answer: (a) There are eight corners in a cube so, total charge for the cube is \(\frac{q}{8}\). Thus, electric flux at \(A=\frac{q}{8 \varepsilon_0}\) [Use of symmetry consideration may be useful in problems of flux calculation. We can imagine the charged particle is placed at the centre of a cube of side 2a . We can observe that the charge is being shared equally by 8 cubes.]
(b) When the charge \(q\) is place at \(B\), middle point of an edge of the cube, it is being shared equally by 4 cubes. Therefore, total flux through the faces of the given cube \(=q / 4 \varepsilon_0\).
(c) When the charge \(q\) is placed at \(C\), the centre of a face of the cube, it is being shared equally by 2 cubes. Therefore, total flux through the faces of the given cube \(=q / 2 \varepsilon_0\).
(d) Similarly, when charge \(q\) is placed at \(Q\), the mid-point of \(B\) and \(C\), it is being shared equally by 2 cubes. Therefore, total flux through the faces of the given cube \(=q / 2 \varepsilon_0\).

Short Questions

Q1.7: A paisa coin is made up of \(\mathrm{Al}-\mathrm{Mg}\) alloy and weighs 0.75 g . It has a square shape and its diagonal measures 17 mm . It is electrically neutral and contains equal amounts of positive and negative charges. Treating the paisa coins made up of only Al, find the magnitude of equal number of positive and negative charges. What conclusion do you draw from this magnitude?

Answer: 1 Molar mass \(M\) of Al has \(N_A=6.023 \times 10^{23}\) atoms
\(\therefore \quad m=\) mass of Al paisa coin has \(N=\frac{N_A}{M} \times m\) atoms
\(\therefore \quad\) Number of aluminium atoms in one paisa coin,
\(
N=\frac{6.023 \times 10^{23}}{26.9815 \mathrm{ab}^5} \times 0.75=1.6742 \times 10^{22}
\)
As charge number of Al is 13 , each atom of Al contains 13 protons and 13 electrons.
\(\therefore\) Magnitude of positive and negative charges in one paisa coin \(=N Z e\)
\(
\begin{aligned}
& =1.6742 \times 10^{22} \times 13 \times 1.60 \times 10^{-19} \mathrm{C} \\
& =3.48 \times 10^4 \mathrm{C}=34.8 \mathrm{kC}
\end{aligned}
\)
This is an enormous amount of charge. Thus, we can conclude that ordinary neutral matter contains large amount of positive and negative charges.

Q1.8: Consider a coin of Example 1.7. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude 34.8 kC . Suppose that these equal charges were concentrated in two point charges seperated by
(i) 1 cm \(\left(\sim \frac{1}{2} \times\right.\) diagonal of the one paisa coin \()\),
(ii) 100 m ( length of a long building), and
(iii) \(10^6 \mathrm{~m}\) (radius of the earth). Find the force on each such point charge in each of the three cases. What do you conclude from these results?

Answer: We know force between two point charges separated at a distance
\(
F=\frac{|q|^2}{4 \pi \varepsilon_0 r^2}
\)
Here, \(\quad q= \pm 34.8 \mathrm{kC}= \pm 3.48 \times 10^4 \mathrm{C}\)
\(
r_1=1 \mathrm{~cm}=10^{-2} \mathrm{~m}, r_2=100 \mathrm{~m}, r_3=10^6 \mathrm{~m}
\)
(i) \(F_1=\frac{|q|^2}{4 \pi \varepsilon_0 r_1^2}=\frac{9 \times 10^9\left(3.48 \times 10^4\right)^2}{\left(10^{-2}\right)^2}=1.09 \times 10^{23} \mathrm{~N}\)
(ii) \(F_2=\frac{|q|^2}{4 \pi \varepsilon_0 r_2^2}=\frac{9 \times 10^9\left(3.48 \times 10^4\right)^2}{(100)^2}=1.09 \times 10^{15} \mathrm{~N}\)
(iii) \(F_3=\frac{|q|^2}{4 \pi \varepsilon_0 r_3^2}=\frac{9 \times 10^9\left(3.48 \times 10^4\right)^2}{\left(10^6\right)^2}=1.09 \times 10^7 \mathrm{~N}\)

Conclusion: Here we can observe that when positive and negative charges in ordinary neutral matter are separated as point charges, they exert very large force. It means, it is very difficult to disturb electrical neutrality of matter.

Q1.9: Fig. 1.10 represents a crystal unit of cesium chloride, \(\mathrm{CsCl}\). The cesium atoms, represented by open circles are situated at the corners of a cube of side 0.40 nm , whereas a \(\mathrm{Cl}\) atom is situated at the centre of the cube. The \(\mathrm{Cs}\) atoms are deficient in one electron while the \(\mathrm{Cl}\) atom carries an excess electron.
(i) What is the net electric field on the \(\mathrm{Cl}\) atom due to eight \(\mathrm{Cs}\) atoms?
(ii) Suppose that the Cs atom at the corner A is missing. What is the net force now on the \(\mathrm{Cl}\) atom due to seven remaining \(\mathrm{Cs}\) atoms?

Answer: (i) The cesium atoms, are situated at the comers of a cube and Cl atom is situated at the centre of the cube. From the given figure, we can analyse that the chlorine atom is at equal distance from all the eight comers of cube where cesium atoms are placed. Thus, due to symmetry the electric field due to all Cs atoms, on Cl atom will cancel out. Hence net electric field at the centre of cube is zero.
\(
\begin{array}{ll}
\text { Hence, } & E=\frac{F}{q} \text { where } F=0 \\
\therefore & E=0
\end{array}
\)

(ii) We define force on a charge particle due to external electric field as \(F=q E\). If eight cesium atoms, are situated at the comers of a cube, the net force on Cl atom is situated at the centre of the cube will be zero as net electric field at the centre of cube is zero. We can write that the vector sum of electric field due to charge \(A\) and electric field due to other seven charges at the centre of cube should be zero or, \(E_A+E_{\text {seven }}\) charges \(=0\)
Hence \(\vec{E}_{\text {seven charges }}=-\vec{E}_A\) or \(\left|\vec{E}_{\text {seven charges }}\right|=\left|\vec{E}_A\right|=\frac{e}{4 \pi \varepsilon_0 r^2}\)
Thus, net force on Cl atom at \(A\) would be,
\(
F=e E_{\text {sevencharges }}=\frac{e^2}{4 \pi \varepsilon_0 r^2}
\)
where, \(r=\) distance between Cl ion and Cs ion.
\(
\begin{aligned}
\text { Here, } r & =\sqrt{(0.20)^2+(0.20)^2+(0.20)^2} \times 10^{-9} \mathrm{~m} \\
& =0.346 \times 10^{-9} \mathrm{~m} \\
\text { Now, } F & =\left(8.99 \times 10^9\right) \frac{\left(1.6 \times 10^{-16}\right)^2}{\left(0.346 \times 10^{-9}\right)^2} \\
& =1.92 \times 10^{-9} \mathrm{~N}, \text { directed from A to } \mathrm{Cl}^{-}
\end{aligned}
\)

Q1.10: Two charges \(q\) and \(-3 q\) are placed fixed on \(x\)-axis separated by distance ‘ \(d\) ‘. Where should a third charge \(2 q\) be placed such that it will not experience any force?

Answer: The force on any charge will be zero only if net electric field at the position of charge is zero. Let electric field is zero at a distance \(x\) from charge \(q\).

\(
\begin{aligned}
& \text { At point } P, \vec{E}_A+\vec{E}_B=0 \Rightarrow\left|\vec{E}_A\right|=\left|\vec{E}_B\right| \\
& \Rightarrow \quad \frac{q}{4 \pi \varepsilon_0 x^2}=\frac{3 q}{4 \pi \varepsilon_0(x+d)^2} \\
& \Rightarrow \quad(x+d)^2=3 x^2 \\
& \Rightarrow \quad x^2+d^2+2 x d=3 x^2 \\
& \therefore \quad 2 x^2-2 d x-d^2=0 \\
& \text { or } \quad x=\frac{d}{2} \pm \frac{\sqrt{3} d}{2}
\end{aligned}
\)
(Negative sign lies between \(q\) and \(-3 q\) and hence is unadaptable.)
Hence \(x=-\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3})\).
Hence if charge \(2 q\) is placed at a distance \(\frac{d}{2}(1+\sqrt{3})\) to the left of \(q\).

Q1.11: Fig. 1.11 shows the electric field lines around three point charges A, B, and C.

(a) Which charges are positive?
(b) Which charge has the largest magnitude? Why?
(c) In which region or regions of the picture could the electric field be zero? Justify your answer.
(i) near A, (ii) near B, (iii) near C, (iv) nowhere.

Answer:

The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In the following figure electric lines of force are originating from \(A\) and terminating at \(B\), hence \(Q_A\) is positive while \(Q_B\) is negative. Also number of electric lines at force linked with \(Q_A\) are more than those linked with \(Q_B\). hence \(\left|Q_A\right|>\left|Q_B\right|\). The electric lines of forces always starts from a positive charge and ends at a negative charge. In case of a single isolated charge, electric lines of force start from positive charge ends at infinity. There is no neutral point between unlike charges. Point between two like charges where electrostatic force is zero is called neutral point. A neutral point may exist between two like charges. Also between two like charges the neutral point is closer to the charge with smaller magnitude.
(i) Here, in the figure, the electric lines of force starts from A and C. Therefore, charges A and C must be positive.
(i) The number of electric lines of forces starting from charge C are maximum, so C must have the largest magnitude.
(iii) From the figure we see that a neutral point exists between charges A and C . Here, more number of electric lines of forces shows higher strength of charge C than A. Thus, electric field is zero near charge A hence neutral point lies near \(A\).

Q1.12: Five charges, \(q\) each are placed at the corners of a regular pentagon of side ‘ \(a\) ‘ (Fig. 1.12).
(a) (i) What will be the electric field at O, the centre of the pentagon?
(ii) What will be the electric field at \(O\) if the charge from one of the corners (say A) is removed?
(iii) What will be the electric field at O if the charge \(q\) at A is replaced by \(-q\)?
(b) How would your answer to (a) be affected if the pentagon is replaced by \(n\)-sided regular polygon with charge \(q\) at each of its corners?

Answer: (i) The point O, the centre of the pentagon is equidistant from all the charges at the endpoint of pentagon. Thus, due to symmetry, the electric field due to all the charges are cancelled out. As a result electric field at O is zero.
(ii) We can write that the vector sum of electric field due to charge \(A\) and electric field due to other four charges at the centre of cube should be zero or,
\(
\vec{E}_A+\vec{E}_{\text {fourcharges }}=0
\)
Hence \(\vec{E}_{\text {fourcharges }}=-\vec{E}_A\) or \(\Rightarrow\left|\vec{E}_{\text {four charges }}\right|=\left|\vec{E}_A\right|\)
When charge \(q\) is removed from \(A\), net electric field at the centre due to remaining charges \(\left|\vec{E}_{\text {four charges }}\right|=\left|\vec{E}_A\right|=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}\) along \(O A\).
(iii) If charge \(q\) at \(A\) is replaced by \(-q\), then electric field due to this negative charge
\(
\vec{E}_{-q}=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} \text { along } O A
\)
Hence net electric field at the centre
\(
\begin{aligned}
& \vec{E}_{\text {net }}=\vec{E}_{-q}+\vec{E}_{\text {fourcharges }}=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}+\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} \\
& \vec{E}_{\text {net }}=\frac{1}{4 \pi \varepsilon_0}-\frac{2 q}{r^2} \text { along } O A
\end{aligned}
\)
(b) If pentagon is replaced by n -sided regular polygon with charge q at each of its comers. Here again, charges are symmetrical about the centre. The net electric field at O would continue to be zero, it doesn’t depend on the number of sides or the number of charges.

Long Answer Type

Q1.13: In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density \(N\), which is maintained a constant. Let the charge on the proton be: \(e_p=-(1+y) e\) where \(e\) is the electronic charge.
(a) Find the critical value of \(y\) such that expansion may start.
(b) Show that the velocity of expansion is proportional to the distance from the centre.

Answer: (a) Let the Universe have a radius R. Assume that the hydrogen atoms are uniformly distributed.
The expansion of the universe will start if the coulomb repulsion on a hydrogen atom, at R is larger than the gravitational attraction.
The hydrogen atom contains one proton and one electron, charge on each hydrogen atom.
\(
e_H=e_P+e=-(1+y) e+e=-y e=|y e|
\)
Let \(E\) be electric field intensity at distance \(R\), on the surface of the sphere, then according to Gauss’ theorem,
\(
\oint \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{S}}=\frac{q_{\text {enclosed }}}{\varepsilon_0}
\)
\(
\begin{array}{ll}
\Rightarrow & E\left(4 \pi R^2\right)=\frac{4}{3} \frac{\pi R^3 N|y e|}{\varepsilon_0} \\
\Rightarrow & E=\frac{1}{3} \frac{N|y e| R}{\varepsilon_0} \dots(i)
\end{array}
\)
Let us suppose the mass of each hydrogen atom \(\approx m_P=\) Mass of a proton and \(G_R=\) gravitational field at distance \(R\) on the sphere.
\(
\begin{array}{ll}
\text { Then } & -4 \pi R^2 G_R=4 \pi G m_P\left(\frac{4}{3} \pi R^3\right) N \\
\Rightarrow & G_R=\frac{-4}{3} \pi G m_P N R \dots(ii)
\end{array}
\)
\(\therefore\) Gravitational force on this atom is
\(
F_G=m_P \times G_R=\frac{-4 \pi}{3} G m_P^2 N R \dots(iii)
\)
Coulomb force on hydrogen atom at \(R\) is
\(
F_C=(y e) E=\frac{1}{3} \frac{N y^2 e^2 R}{\varepsilon_0} \quad \text { [from Eq. (i)] }
\)
Now, to start expansion \(F_C>F_G\) and critical value of \(Y\) to start expansion would be when
\(
\begin{aligned}
& F_C=F_G \\
\Rightarrow \quad & \frac{1}{3} \frac{N y^2 e^2 R}{\varepsilon_0}=\frac{4 \pi}{3} G m_P^2 N R \\
\Rightarrow \quad & y^2=\left(4 \pi \varepsilon_0\right) G\left(\frac{m_P}{e}\right)^2
\end{aligned}
\)
\(
\begin{aligned}
& =\frac{1}{9 \times 10^9} \times\left(6.67 \times 10^{-11}\right) \\
& =\frac{1}{9 \times 10^9} \times\left(6.67 \times 10^{-11}\right)\left(\frac{\left(1.66 \times 10^{-27}\right)^2}{\left(1.6 \times 10^{-19}\right)^2}\right)=79.8 \times 10^{-38} \\
\Rightarrow \quad y & =\sqrt{79.8 \times 10^{-38}}=8.9 \times 10^{-19} \simeq 10^{-18}
\end{aligned}
\)
Hence \(10^{-18}\) is the required critical value of \(y\) corresponding to which expansion of universe would start.

(b) Net force experienced by the hydrogen atom is given by
\(
F=F_C-F_G=\frac{1}{3} \frac{N y^2 e^2 R}{\varepsilon_0}-\frac{4 \pi}{3} G m_P^2 N R
\)
Because of this net force, the hydrogen atom experiences an acceleration such that
\(
\begin{aligned}
m_P \frac{d^2 R}{d t^2} & =F=\frac{1}{3} \frac{N y^2 e^2 R}{\varepsilon_0}-\frac{4 \pi}{3} G m_P^2 N R \\
& =\left(\frac{1}{3} \frac{N y^2 e^2}{\varepsilon_0}-\frac{4 \pi}{3} G m_P^2 N\right) R
\end{aligned}
\)
\(
\frac{d^2 R}{d t^2}=\frac{1}{m_P}\left[\frac{1}{3} \frac{N y^2 e^2}{\varepsilon_0}-\frac{4 \pi}{3} G m_P^2 N\right]
\)
\(
\Rightarrow \quad \frac{d^2 R}{d t^2}=\alpha^2 R \dots(iv)
\)
where, \(\alpha^2=\frac{1}{m_P}\left[\frac{1}{3} \frac{N Y^2 e^2}{\varepsilon_0}-\frac{4 \pi}{3} G m_P^2 N\right]\)
The solution of Eq. (iv) is given by \(R=A e^{\alpha t}+B e^{-\alpha t}\). We are looking for expansion, here, so \(B=0\) and \(R=A e^{\alpha t}\).
\(\Rightarrow \quad\) Velocity of expansion,
\(
v=\frac{d R}{d t}=A e^{\alpha t}(\alpha)=\alpha A e^{\alpha t}=\alpha R
\)
Hence, \(v \propto R\), i.e., velocity of expansion is proportional to the distance from the centre.

Q1.14: Consider a sphere of radius R with charge density distributed as
\(
\begin{array}{rlrl}
\rho(r) & =k r & \text { for } r \leq R \\
& =0 & & \text { for } r>R .
\end{array}
\)
(a) Find the electric field at all points \(r\).
(b) Suppose the total charge on the sphere is \(2 e\) where \(e\) is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.

Answer:  (a) The expression of charge density distribution in the sphere suggests that the electric field is radial.

Let us consider a sphere \(S\) of radius \(R\) and two hypothetic spheres of radius \(r<R\) and \(r>R\).
Let us first consider for point \(r<R\), electric field intensity will be given by,
\(
\oint \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{S}}=\frac{1}{\varepsilon_0} \int \rho d V
\)
Here \(d V=4 \pi r^2 d r\)
\(
\begin{aligned}
& \Rightarrow \quad \oint \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{S}}=\frac{1}{\varepsilon_0} 4 \pi K \int_0^r r^3 d r(\because \rho(r)=K r) \\
& \Rightarrow \quad(E) 4 \pi r^2=\frac{4 \pi K}{\varepsilon_0} \frac{r^4}{4} \\
& \text { We get, } E=\frac{1}{4 \varepsilon_0} K r^2
\end{aligned}
\)
As charge density is positive, it means the direction of \(\mathbf{E}\) is radially outwards.
Now consider points \(r>R\), electric field intensity will be given by
\(
\begin{aligned}
& \oint \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \mathbf{S}=\frac{1}{\varepsilon_0} \int \rho d V \\
\Rightarrow \quad & E\left(4 \pi r^2\right)=\frac{4 \pi K}{\varepsilon_0} \int_0^R r^3 d r=\frac{4 \pi K}{\varepsilon_0} \frac{R^4}{4}
\end{aligned}
\)
which gives, \(E=\frac{K}{4 \varepsilon_0} \frac{R^4}{r^2}\)
Here also the charge density is again positive. So, the direction of \(\mathbf{E}\) is radially outward.

(b)

The two protons must be placed symmetrically on the opposite sides of the centre along a diameter. This can be shown by the figure given below. Charge on the sphere,
\(
\begin{aligned}
q & =\int_0^R \rho d V=\int_0^R(K r) 4 \pi r^2 d r \\
q & =4 \pi K \frac{R^4}{4}=2 e \\
\because \quad K & =\frac{2 e}{\pi R^4}
\end{aligned}
\)
The protons 1 and 2 are embedded at distance \(r\) from the centre of the sphere as shown, then attractive force on proton 1 due to charge distribution is
\(
F_1=e E=\frac{-e K r^2}{4 \varepsilon_0}
\)
And repulsive force on proton 1 due to proton 2 is
\(
F_2=\frac{e^2}{4 \pi \varepsilon_0(2 r)^2}
\)
Net force on proton \(1, F=F_1+F_2\)
\(
F=\frac{-e K r^2}{4 \varepsilon_0}+\frac{e^2}{16 \pi \varepsilon_0 r^2}
\)
So, \(\quad F=\left[\frac{-e r^2}{4 \varepsilon_0} \frac{Z e}{\pi R^4}+\frac{e^2}{16 \pi \varepsilon_0 r^4}\right]\)
Thus, net force on proton 1 will be zero, when
\(
\frac{e r^2 2 e}{4 \varepsilon_0 \pi R^4}=\frac{e^2}{16 \pi \varepsilon_0 r}
\)
\(
\Rightarrow \quad r^4=\frac{R^4}{8}
\)
Hence the protons must be at a distance \(r=\frac{R}{(8)^{1 / 4}}\) from the centre.

Q1.15: Two fixed, identical conducting plates \((\alpha \& \beta)\), each of surface area \(S\) are charged to \(-Q\) and \(q\), respectively, where \(Q>q>0\). A third identical plate \((\gamma)\), free to move is located on the other side of the plate with charge \(q\) at a distance \(d\) (Fig 1.13). The third plate is released and collides with the plate \(\beta\). Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst \(\beta  \&  \gamma\).
(a) Find the electric field acting on the plate \(\gamma\) before collision.
(b) Find the charges on \(\beta\) and \(\gamma\) after the collision.
(c) Find the velocity of the plate \(\gamma\) after the collision and at a distance \(d\) from the plate \(\beta\).

Answer: (a) Net electric field at plate \(\gamma\) before collision is vector sum of electric field at plate \(\gamma\) due to plate \(\alpha\) and \(\beta\).
The electric field at plate \(\gamma\) due to plate \(\alpha\) is \(\vec{E}_1=\frac{Q}{S\left(2 \varepsilon_0\right)}(-\hat{i})\),
The electric field at plate \(\gamma\) due to plate \(\beta\) is \(\vec{E}_2=\frac{q}{S\left(2 \varepsilon_0\right)}(\hat{i})\),
Hence, the net electric field at plate \(\gamma\) before collision is
\(
\begin{gathered}
\vec{E}=\vec{E}_1+\vec{E}_2=\frac{q-Q}{S\left(2 \varepsilon_0\right)}(\hat{i}) \\
\vec{E}=\vec{E}_1+\vec{E}_2=\frac{Q-q}{S\left(2 \varepsilon_0\right)}(-\hat{i})
\end{gathered}
\)
or \(\quad \frac{Q-q}{S\left(2 \varepsilon_0\right)}\) to the left, if \(Q>q\)
(b) During collision, plates \(\beta\) and \(\gamma\) are in contact with each other, hence their potentials become same.
Suppose charge on plate \(\beta\) is \(q_1\) and charge on plate \(\gamma\) is \(q_2\). At any point \(O\), in between the two plates, the electric field must be zero.
Electric field at \(O\) due to plate \(\alpha, \vec{E}_\alpha=\frac{Q}{S\left(2 \varepsilon_0\right)}(-\hat{i})\)
Electric field at \(O\) due to plate \(\beta, \vec{E}_2=\frac{q_1}{S\left(2 \varepsilon_0\right)}(\hat{i})\)
Electric field at \(O\) due to plate \(\gamma, \vec{E}_\gamma=\frac{q_2}{S\left(2 \varepsilon_0\right)}(-\hat{i})\)
As the electric field at \(O\) is zero, therefore
\(
\begin{array}{ll}
& \frac{Q+q_2}{S\left(2 \varepsilon_0\right)}=\frac{q_1}{S\left(2 \varepsilon_0\right)} \\
\because \quad Q+q_2=q_1 \dots(i)
\end{array}
\)
As there is no loss of charge on collision,
\(
Q+q=q_1+q_2 \dots(ii)
\)
On solving Eqs. (i) and (ii), we get
\(
\begin{aligned}
& q_1=(Q+q / 2)=\text { charge on plate } \beta \\
& q_1=(q / 2)=\text { charge on plate } \gamma
\end{aligned}
\)
(c) Let the velocity be \(v\) at the distance \(d\) from plate \(\beta\) after the collision. If \(m\) is the mass of the plate \(\gamma\), then the gain in K.E. over the round trip must be equal to the work done by the electric field.
After the collision, electric field at plate \(\gamma\) is
\(
\vec{E}_2=\frac{Q}{2 \varepsilon_0 S}(-\hat{i})+\frac{(Q+q / 2)}{2 \varepsilon_0 S} \hat{i}=\frac{q / 2}{2 \varepsilon_0 S} \hat{i}
\)
Just before collision, electric field at plate \(\gamma\) is \(\vec{E}_1=\frac{Q-q}{2 \varepsilon_0 S} \hat{i}\).
If \(F_1\) is force on plate \(\gamma\) before collision, then \(\vec{F}_1=\vec{E}_1 Q=\frac{(Q-q) Q}{2 \varepsilon_0 S} \hat{i}\). And \(\quad \vec{F}_2=\vec{E}_2 \frac{q}{2}=\frac{(q / 2)^2}{2 \varepsilon_0 S} \hat{i}\)
Total work done by the electric field is round trip movement of plate \(\gamma\),
\(
\begin{aligned}
W & =\left(F_1+F_2\right) d \\
& =\frac{\left[(Q-q) Q+(q / 2)^2\right] d}{2 \varepsilon_0 S}=\frac{(Q-q / 2)^2 d}{2 \varepsilon_0 S}
\end{aligned}
\)
If \(m\) is the mass of plate \(\gamma\), the KE gained by the plate \(=\frac{1}{2} m v^2\)
According to work-energy principle, \(\frac{1}{2} m v^2=W \Rightarrow \frac{1}{2} m v^2=\frac{(Q-q / 2)^2 d}{2 \varepsilon_0 S}\)
\(
\Rightarrow \quad v=(Q-q / 2)\left(\frac{d}{m \varepsilon_0 S}\right)^{1 / 2}
\)

Q1.16: There is another useful system of units, besides the SI/mks A system, called the cgs (centimeter-gram-second) system. In this system Coloumb’s law is given by
\(
\mathbf{F}=\frac{\mathrm{Qq}}{r^2} \hat{\boldsymbol{r}}
\)
where the distance \(r\) is measured in \(\mathrm{cm}\left(=10^{-2} \mathrm{~m}\right), F\) in dynes \(\left(=10^{-5} \mathrm{~N}\right)\) and the charges in electrostatic units (es units), where
\(
\text { l es unit of charge }=\frac{1}{[3]} \times 10^{-9} \mathrm{C}
\)
The number [3] actually arises from the speed of light in vaccum which is now taken to be exactly given by \(c=2.99792458 \times 10^8\) \(\mathrm{m} / \mathrm{s}\). An approximate value of \(c\) then is \(c=[3] \times 10^8 \mathrm{~m} / \mathrm{s}\).
(i) Show that the coloumb law in cgs units yields 1 esu of charge \(=1(\text { dyne })^{1 / 2} \mathrm{~cm}\).
Obtain the dimensions of units of charge in terms of mass \(M\), length \(L\) and time \(T\). Show that it is given in terms of fractional powers of \(M\) and \(L\).
(ii) Write 1 esu of charge \(=x \mathrm{C}\), where \(x\) is a dimensionless number. Show that this gives
\(
\frac{1}{4 \pi \epsilon_0}=\frac{10^{-9}}{x^2} \frac{\mathrm{~N} \cdot \mathrm{~m}^2}{\mathrm{C}^2}
\)
With \(x=\frac{1}{[3]} \times 10^{-9}\), we have
\(
\begin{aligned}
& \frac{1}{4 \pi \epsilon_0}=[3]^2 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2} \\
& \text { or, } \frac{1}{4 \pi \epsilon_0}=(2.99792458)^2 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2} \text { (exactly). }
\end{aligned}
\)

Answer: (i) According to the relation, \(F=\frac{Q q}{r^2}=1\) dyne \(=\frac{[1 \mathrm{esu} \text { of charge }]^2}{[1 \mathrm{~cm}]^2}\)
So, 1 esu of charge \(=(1 \text { dyne })^{1 / 2} \times 1 \mathrm{~cm}\)
\(
=F^{1 / 2} \cdot L=\left[M L T^{-2}\right]^{1 / 2}[L]=\left[M^{1 / 2} L^{3 / 2} T^{-1}\right]
\)
\(\Rightarrow\) Hence, \([1\) esu of charge \(]=M^{1 / 2} L^{3 / 2} T^{-1}\)
Thus the charge in C.G.S unit (in esu) is represented in terms of fractional powers \(\frac{1}{2}\) of \(M\) and \(\frac{3}{2}\) of \(L\).
(ii) If two charges each of magnitude 1 esu separated by 1 cm , the Coulomb force on the charges is 1 dyne \(=10^{-5} \mathrm{~N}\).
Let 1 esu of charge \(=x C\), where \(x\) is a dimensionless number. We can consider a situation, two charges of magnitude \(x C\) separated by \(10^{-2} \mathrm{~m}\). The force between the charges
\(
\begin{aligned}
& F=\frac{1}{4 \pi \varepsilon_0} \frac{x^2}{\left(10^{-2}\right)^2}=1 \text { dyne }=10^{-5} \mathrm{~N} \\
\therefore \quad & \frac{1}{4 \pi \varepsilon_0}=\frac{10^{-9}}{x^2} \frac{\mathrm{Nm}^2}{\mathrm{C}^2}
\end{aligned}
\)
Taking, \(\quad x=\frac{1}{|3| \times 10^9}\),
we get \(\quad \frac{1}{4 \pi \varepsilon_0}=10^{-9} \times|3|^2 \times 10^{18}\)
\(
\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}
\)
If \(|3| \rightarrow 2.99792458\),
we get \(\frac{1}{4 \pi \varepsilon_0}=8.98755 \ldots \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}\) exactly.

Q1.17: Two charges \(-q\) each are fixed separated by distance \(2 d\). A third charge \(q\) of mass \(m\) placed at the mid-point is displaced slightly by \(x(x \ll d)\) perpendicular to the line joining the two fixed charged as shown in Fig. 1.14. Show that \(q\) will perform simple harmonic oscillation of time period.
\(
T=\left[\frac{8 \pi^3 \varepsilon_0 m d^3}{q^2}\right]^{1 / 2}
\)

Answer: 

Let the charge \(q\) is displaced slightly by \(x(x \ll d)\) perpendicular to the line joining the two fixed charges. Net force on the charge \(q\) will be towards \(O\). The motion of charge \(q\) to be simple harmonic, if the force on charge \(q\) must be proportional to its distance from the centre \(O\) and is directed towards \(O\).
Net force on the charge \(F_{\text {net }}=2 F \cos \theta\)
Here \(F=\frac{1}{4 \pi \varepsilon_0} \frac{q(q)}{r^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{\left(d^2+x^2\right)}\)
And \(\cos \theta=\frac{x}{\sqrt{x^2+d^2}}\)
Hence, \(F_{\mathrm{net}}=2\left[\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{\left(d^2+x^2\right)}\right]\left[\frac{x}{\sqrt{x^2+d^2}}\right]\)
\(
=\frac{1}{2 \pi \varepsilon_0} \frac{q^2 x}{\left(d^2+x^2\right)^{3 / 2}}=\frac{1}{2 \pi \varepsilon_0} \frac{q^2 x}{d^3\left(1+\frac{x^2}{d^2}\right)^{3 / 2}}
\)
As \(x \ll d\), then \(F_{\text {net }}=\frac{1}{2 \pi \varepsilon_0} \frac{q^2 x}{d^3}\) or \(F_{\text {net }}=K x\)
i.e., force on charge \(q\) is proportional to its displacement from the centre \(O\) and it is directed towards \(O\). Hence, motion of charge \(q\) would be simple harmonic, where \(\omega=\sqrt{\frac{K}{m}}\)
and \(T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{K}}\)
\(
\Rightarrow \quad T=2 \pi \sqrt{\frac{m \cdot 4 \pi \varepsilon_0 d^3}{2 q^2}}=\left[\frac{8 \pi^3 \varepsilon_0 m d^3}{q^2}\right]^{1 / 2}
\)

Q1.18: Total charge \(-Q\) is uniformly spread along length of a ring of radius \(R\). A small test charge \(+q\) of mass \(m\) is kept at the centre of the ring and is given a gentle push along the axis of the ring.
(a) Show that the particle executes a simple harmonic oscillation.
(b) Obtain its time period.

Answer: Let the charge \(q\) is displaced slightly by \(z(z \ll R)\) along the axis of ring. Let force on the charge \(q\) will be towards \(O\). The motion of charge \(q\) to be simple harmonic, if the force on charge \(q\) must be proportional to \(z\) and is directed towards \(O\).

A gentle push on \(q\) along the axis of the ring gives rise to the situation shown in the figure below.

Electric field at axis of the ring at a distance \(z\) from the centre of ring
\(
\begin{aligned}
&E=\frac{1}{4 \pi \varepsilon_0} \frac{Q z}{\left(R^2+z^2\right)^{3 / 2}} ; \text { towards } O\\
&\text { Net force on the charge } F_{\text {net }}=q E\\
&\begin{aligned}
F_{\text {net }} & =\frac{1}{4 \pi \varepsilon_0} \frac{q Q z}{\left(R^2+z^2\right)^{3 / 2}} \\
\Rightarrow \quad F_{\text {net }} & =\frac{1}{4 \pi \varepsilon_0} \frac{q Q z}{R^3\left(\frac{1+z^2}{R^2}\right)^{3 / 2}}
\end{aligned}
\end{aligned}
\)
\(
\text { As } z \ll R \text { then, } F_{\text {net }} \equiv \frac{1}{4 \pi \varepsilon_0} \frac{q Q z}{R^3}
\)
or \(\quad \vec{F}_{\mathrm{net}}=-K \vec{z}\)
where \(K=\frac{Q q}{4 \pi \varepsilon_0 R^3}=\) constant
Clearly, force on \(q\) is proportional to negative of its displacement. Therefore, motion of \(q\) is simple harmonic.
\(
\begin{aligned}
\omega & =\sqrt{\frac{K}{m}} \text { and } T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{K}} \\
T & =2 \pi \sqrt{\frac{m 4 \pi \varepsilon_0 R^3}{Q q}} \\
\Rightarrow \quad T & =2 \pi \sqrt{\frac{4 \pi \varepsilon_0 m R^3}{Q q}}
\end{aligned}
\)