NCERT Exemplar Q & A

Hint

(a) Key concept: Total force acting on a given charge due to the number of charges is the vector sum of the individual forces acting on that charge due to all the charges.
Consider the number of charges \(Q_1, Q_2, Q_3, \ldots\) are applying force on a charge \(Q\).
Net force on \(Q\) will be
\(
\vec{F}_{\mathrm{net}}=\vec{F}_1+\vec{F}_2+\ldots+\vec{F}_{n-1}+\vec{F}_n
\)

We know that like charges repel and unlike charges attract. The net electrostatic force on the charge qx by the charges \(\mathrm{q}_2\) and \(\mathrm{q}_3\) is along the positive x -direction. Hence the nature of force between \(\mathrm{q}_1, \mathrm{q}_2\) and \(\mathrm{q}_1, \mathrm{q}_3\) should be attractive. It means \(q x\) should be negative. This can be represented by the figure given alongside: Now a positive charge Q is placed at \((\mathrm{x}, 0)\), hence the nature of force between \(\mathrm{q}_1\) and Q (positive) will be attractive and the force on \(q_1\) by the charge \(Q\) should be along positive \(x\)-axis direction. Now we can say that net force on the charge qx due to charges \(\mathrm{q}_2, \mathrm{q}_3\) and Q should be along the same direction as given in the diagram alongside:
Now it is clear from the figure given above that the force on \(q \times\) shall increase along the positive \(x\)-axis due to the presence of positive charge Q placed at \((\mathrm{x}, 0)\).

Hint

Key concept:
Electric field lines come out of positive charge and go into the negative charge.
Tangent to the field line at any point gives the direction of the field at that point.
Field lines are always normal to the conducting surface.
Field lines do not exist inside a conductor.

The explanation to this problem-can be done by keeping two things in mind.
(i) Concept of induction
(ii) The electric field lines interact with a conducting body normally.

Let us discuss the phenomenon of induction involved in this case. When a positive point charge is brought near an isolated conducting sphere without touching the sphere, then the free electrons in the sphere are attracted towards the positive charge. Thus, the left surface of sphere has an excess of negative charge and the right surface of sphere has an excess of positive charge. It should be noted that both kinds of charges are bound in the metal sphere and cannot escape. They, therefore, reside on the surface of the sphere.

An electric field lines start from a positive point charge and ends at negative charge induced on the left surface of sphere. Also, electric field line emerges from a positive charge, in case of single charge and ends at infinity.
Here, all these conditions are fulfilled in Fig. (i).

Hint

(d) Key concept: According to Gauss’ law of electrostatics, the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity,
\(
\text { i.e., } \phi=\frac{Q_{\text {enclosed }}}{\varepsilon_0}
\)
Thus, electric flux through a surface doesn’t depend on the shape, size or area of a surface but it depends on the amount of charge enclosed by the surface.
In given figures the charge enclosed are same that means the electric flux through all the surfaces should be the same. Hence option (d) is correct.

Hint

(b) Key concept: According to Gauss’ law, the term \(q_{\text {enclosed }}\) on the right side of the equation
\(
\oint_s \vec{E} \cdot d \vec{S}=\frac{q_{\text {enclosed }}}{\varepsilon_0}
\)
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.
Hence in given question, E on LHS of the above equation will have a contribution from all charges while q on the RHS will have a contribution from \(\mathrm{q}_2\) and \(\mathrm{q}_4\) only. Hence option (b) is correct.

Hint

(c) Key concept: If the lines of forces are equidistant and parallel straight lines, the field is uniform and if either lines of force are not equidistant, or straight line or both, the field will be non-uniform. The number of electric field Imes passing per unit area is proportional to the strength of electric field. For example, see the following figures:

Hence in given question, from given pattern of electric field lines it is clear that the strength of electric field decreases from left to right. As a result force on charges also decreases from left to right. Here in given figure, the force on charge \(-q\) is greater than force on charge \(+q\) in turn dipole will experience a force towards left. Hence option (c) is correct.

Hint

(a) If a point positive charge is placed near an isolated conducting plane, free electrons are attracted towards the positive charge. Result of this some negative charge develops on the surface of the plane towards the positive charge side and an equal positive charge develops on opposite side of the plane. The electric field lines are away from positive charge and perpendicular to the surface. Hence the field at a point \(P\) on the other side of the plane is directed perpendicular to the plane and away from the plane, hence option (a) is correct.

Hint

(a) In case of a uniformly positive charged hemisphere, if a point situated at a point on a diameter away from the centre, the electric field should be perpendicular to the diameter. In this case the component of electric field intensity parallel to the diameter cancel out.

Hint

(c, d)

\(\oint_S E . d S=0\) represents electric flux over the closed surface.
In general, \(\oint_S\) E.dS means the algebraic sum of number of flux lines entering the surface and number of flux lines leaving the surface.
When \(\oint_S E d S=0\), it means that the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
Now, from Gauss’ law, we know that \(\oint_S E . d S=\frac{q}{\varepsilon_0}\) where \(q\) is charge enclosed by the surface. When \(\int_S E d S=0, q=0\) i.e., net charge enclosed by the surface must be zero. Therefore, all other charges must necessarily be outside the surface. This is because charges outside because of the fact that charges outside the surface do not contribute to the electric flux.

Hint

(b, d) We cannot define electric field at the position of a charge, so we cannot say that electric field is always continuous. Hence option (a) is ruled out and option (d) is the correct choice. The electric field due to any charge will be continuous, if there is no other charge in the medium. It will be discontinuous if there is a charge at the point under consideration, hence option (b) is correct.

Hint

(b, d) From Gauss’ law, we know \(\oint_s \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{S}}=\frac{q}{\varepsilon_0}\), here q is the charge enclosed by the closed surface. If \(\oint_s \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{S}}=0\) then \(q=0\), i.e., net charge enclosed by the surface must be zero.
If the charge is outside the surface, then charge enclosed by the surface is \(q=0\) and thus, \(\oint_s \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{S}}=0\). Hence options (b) and (d) are correct.e the surface \(\oint \vec{E} . d \vec{S}=0\). Hence, Options (b) and (d) are correct.

Hint

(c, d) From Gauss’ law, we know \(\oint_s \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{S}}=\frac{{q}_{\text {enclosed }}}{\varepsilon_0}\). in left side equation. the electric field is due to all the charges present both inside as well as outside the Gaussian surface. Hence if \(q_{\text {enclosed }}=0\), it cannot be said that the electric field is necessarily zero.
If there are various types of charges in a region and total charge is zero, the region may be supposed to contain a number of electric dipoles.
Therefore, at points outside the region (may be anywhere w.r.t. electric dipoles), the dominant electric field \(\propto 1 / r^3[latex] for large [latex]r\).
The electric field is conservative, work done to move a charged particle along a closed path, away from the region will be zero.

Hint

\(
\text { (a, c) From Gauss’ law, we know } \oint_s \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{S}}=\frac{q_{\text {enclosed }}}{\varepsilon_0} \text {. }
\)
Thus, from figure, Total charge inside the Gaussian surface
\(
q_{\text {enclosed }}=Q-2 Q=-Q
\)
The charge 5Q lies outside the surface, thus it makes no contribution to electric flux through the given surface. Hence options (a) and (c) are correct.

Hint

\((a, b, c, d)\) The positive charge \(Q\) is uniformly distributed along the circular ring then electric field at the centre of ring will be zero, hence no force is experienced by the charge if it is placed at the centre of the ring.
Now the charge is displaced away from the centre in the plane of the ring. There will be net electric field opposite to displacement will push back the charge towards the centre of the ring if the charge is positive. If charge is negative, it will experience net force in the direction of displacement and the charge will continue moving till it hits the ring. Also this negative charge is in an unstable equilibrium. Hence options (a), (b), (c) and (d) are correct.

The direction of electric field on the axis of a positively charged ring is along the axis of the ring and away from the centre of ring. If a negative charge is shifted away from the centre along the axis of ring, charge will experience a net force towards the centre and return to the centre and will perform SHM for small displacement along the axis.

Hint

(c) The crowding of electric field lines at a point shows the strength of the field at that point. More the crowding of field lines, more will be the field strength. At points A and C, there’s equal crowding, whereas at point B, the lines are far apart. Therefore, \(E_A=E_C>E_B\)

Note: Electric field intensity at a particular point is directly proportional to the number of electric line of forces crossing per unit area.

Hint

(d) The electric potential energy, \(U\) , between the two charges, \(\mathrm{q}_1\) and \(\mathrm{q}_2\), separated by the distance, \(r\), is given as \(U=k \frac{q_1 q_2}{r}\),
Where \({k}=\) constant
As the distance between the charges is increased, the energy will decrease if both the charges are of similar nature. But if the charges are oppositely charged, the energy will become less negative and, hence, will increase.

Hint

(a) The electric potential energy, U, of a positive charge, q, in a potential, V, is given by \(\mathrm{U}=\mathrm{qV}\). As the charge is moved from a low-potential region to a high-potential region, i.e. as V is increased, U will increase.

Hint

(d) Let the distance between the points A and B be r.
Let us take a point P at a distance x from \({A}({x}<{r})\).
Electric potential V at point P due to two charges of equal magnitude q is given by
\(
\begin{aligned}
& V=\frac{q}{4 \pi \epsilon_0 x}+\frac{q}{4 \pi \epsilon_0(r-x)} \\
& \Rightarrow V=\frac{q r}{4 \pi \epsilon_0 x(r-x)}
\end{aligned}
\)
Now, differentiating V with respect to x , we get
\(
\frac{d V}{d x}=\frac{-q r(r-2 x)}{4 \pi \in_0 x^2(r-x)^2}
\)
It can be observed that \(\frac{d V}{d x}<0\) for \({x}<{r} / 2\).
Thus, the potential is decreasing first.
At \(x=r / 2\), the potential is minimum.
As \(\frac{d V}{d x}>0\) for \({x}>{r} / 2\), the potential is increasing after \({x}={r} / 2\).

Hint

(a) Electric field is along the positive \(x\)-axis.
Circle centered at the origin with radius \(a\).
Points on the circle: \(A(a, 0), B(0, a), C(-a, 0), D(0,-a)\).
Electric potential \(V\) is related to the electric field \(\vec{E}\) by \(V=-\int \vec{E} \cdot d \vec{r}\).
For a uniform electric field, \(V=-\vec{E} \cdot \vec{r}\).
Potential at point A:
The position vector of point A is \(\vec{r}_A=a \hat{i}\).
The electric field is \(\vec{E}=E \hat{i}\).
The potential at A is \(V_A=-\vec{E} \cdot \vec{r}_A=-E \hat{i} \cdot a \hat{i}=-E a\).
Potential at point B:
The position vector of point B is \(\vec{r}_B=a \hat{j}\).
The potential at B is \(V_B=-\vec{E} \cdot \vec{r}_B=-E \hat{i} \cdot a \hat{j}=0\).
Potential at point C:
The position vector of point C is \(\vec{r}_C=-a \bar{i}\).
The potential at C is \(V_C=-\vec{E} \cdot \vec{r}_C=-E \hat{i} \cdot(-a \hat{i})=E a\).
Potential at point D:
The position vector of point D is \(\vec{r}_D=-a \bar{j}\).
The potential at D is \(V_D=-\vec{E} \cdot \vec{r}_D=-E \hat{i} \cdot(-a \hat{j})=0\).
\(
\begin{aligned}
& V_A=-E a \\
& V_B=0 \\
& V_C=E a \\
& V_D=0
\end{aligned}
\)
Since \(E\) and \(a\) are positive, \(V_A\) is the minimum potential.
The potential is minimum at point A.

Hint

(d) If a body is rubbed with another body, it’ll either gain some electrons from the other body and become negatively charged or it’ll lose some electrons to the other body and become positively charged. Since electrons have a very small mass, gaining or losing them can slightly change the body’s overall mass. Gain of electrons increases the weight of a body slightly and loss of electrons reduces the weight slightly.

Hint

(a) Given, The dipole is placed in a uniform electric field.
The dipole consists of two equal and opposite charges.
The force on a charge in an electric field is given by \(F=q E\), where \(q\) is the charge and \(E\) is the electric field strength.
The electric field is uniform, meaning it has the same magnitude and direction at all points in the given region.
The force on the positive charge is \(F_{+}=q E\).
The force is in the direction of the electric field.
The force on the negative charge is \(F_{-}=-q E\).
The force is in the opposite direction of the electric field.
The net force is \(F_{\text {net }}=F_{+}+F_{-}=q E-q E=0\).
The net electric force on the dipole is zero.

Hint

(c) Points \(A, B\), and \(C\) are equidistant from charge \(q\).
\(W_A, W_B\), and \(W_C\) are the work done in moving a point charge from \(P\) to \(A, B\), and \(C\) respectively.
The work done in moving a charge between two points depends only on the potential difference between the points.

Points at the same distance from a charge have the same electric potential. Work done \(W\) is given by \(W=q \Delta V\), where \(q\) is the charge and \(\Delta V\) is the potential difference.
The electric potential \(\boldsymbol{v}\) due to a point charge \(q\) at a distance \(r\) is given by:
\(V=k \frac{q}{r}\)

Since \(A, B\), and \(C\) are at the same distance \(r\) from \(q\), their potentials are equal:
\(V_A=V_B=V_C=k \frac{q}{r}\)
The work done in moving a charge \(q_0\) from \(P\) to \(A\) is:
\(W_A=q_0\left(V_A-V_P\right)\)

Similarly, for \(B\) and \(C\) :
\(W_B=q_0\left(V_B-V_P\right)\)
\(W_C=q_0\left(V_C-V_P\right)\)

Since \(V_A=V_B=V_C\), we have:
\(W_A=W_B=W_C\)
The work done in taking the point charge from \(P\) to \(A, B\), and \(C\) is the same.

Hint

(a)

Given that: A point charge \((Q)\) is rotating along a circle in the electric field generated by another point charge \(P\).
Electric field due to point charge is spherical or it is outside the given circle.
A charge is moving along the circle, so its displacement vector is always perpendicular to the electric field vector.
Now, work done \(W=\vec{F} \cdot \vec{d x}\) and force \(\vec{F}=Q \cdot \vec{E}\)
\(
\Rightarrow W=Q \cdot \vec{E} \cdot \vec{d x}
\)
As displacement vector is perpendicular to Electric field
\(
\therefore \vec{E} \perp \vec{d x}
\)
So that dot product of \(\vec{E}\) and \(\vec{d x}\) is
\(
\vec{E} \cdot \vec{d x}=\vec{E} \cdot \vec{d x} \cos \theta
\)
Where \(\theta=90^{\circ}\)
\(
\therefore \vec{E} \cdot \vec{d x}=\vec{E} \vec{d x} \cos 90^{\circ}=0
\)
Hence work done is zero at all points.

Hint

(a) The total charge of the universe is constant.
Explanation:
Based on the principle of conservation of charge, the total charge in a closed system remains constant. Since the universe can be considered a closed system, this means the total positive charge and total negative charge must also be constant, resulting in a constant overall charge.

Why other options are incorrect:
(b) The total positive charge of the universe is constant:

While the total charge is constant, the distribution of positive and negative charges can shift. This means the total positive charge alone is not necessarily constant. For instance, if a particle with a positive charge interacts with another particle and loses its charge, the total positive charge would decrease.
(c) The total negative charge of the universe is constant:

Similar to (b), the total negative charge is not constant independently. It’s only the sum of positive and negative charges (the total charge) that is constant.
(d) The total number of charged particles in the universe is constant:

The number of charged particles can change through processes like pair creation and annihilation, where particles are created or destroyed. While these processes always involve equal amounts of positive and negative charges, thus maintaining the overall charge constancy, the number of particles themselves is not fixed.

Hint

(c, d) Electric field is a vector quantity. The electric field at a point due to a number of point charges is the vector sum of electric field due to individual charges. So, when a positive charge is brought into an electric field, the electric field due to the positive charge is added to the electric field already present. Therefore, the electric field increases.

When a negative charge is brought into an electric field, the electric field due to the negative charge is subtracted from the electric field already present. Therefore, the electric field decreases.

Hint

(e) Electric field is the negative gradient of electric potential: \(E=-\frac{d V}{d x}\).
Electric potential is a scalar quantity.
Electric field is a vector quantity.
(a) Consider a region of space with a uniform potential.
The electric field is the negative gradient of the potential.
If the potential is uniform (constant), its gradient is zero.
Therefore, \(E=0\) does not imply \(V=0\).
Statement (a) is incorrect.
(b) Consider the midpoint between two equal and opposite charges.
The electric potential at the midpoint is zero.
However, the electric field at the midpoint is non-zero.
Therefore, \(V=0\) does not imply \(E=0\).
Statement (b) is incorrect.
(c) Consider the midpoint between two equal and opposite charges.
The electric field at the midpoint is non-zero.
The electric potential at the midpoint is zero.
Therefore, \(E \neq 0\) does not imply \(\boldsymbol{V} \neq 0\).
Statement (c) is incorrect.
(d) Consider a point near a single point charge.
The electric potential at that point is non-zero.
The electric field at that point is also non-zero.
However, consider a region of space with a uniform non-zero potential.
The electric field in this region is zero.
Therefore, \(V \neq 0\) does not imply \(E \neq 0\).
Statement (d) is incorrect.
None of the statements are correct.

Note: Yes, it’s possible for the electric potential at a point to be zero while the electric field is not. This occurs when dealing with a dipole, where two equal and opposite charges are separated by a distance. At the midpoint, the electric potential is zero, but the electric field is not.

Hint

(b,c)  Electric potential at \(x_1=-1 \mathrm{~cm}: V_1=120 \mathrm{~V}\)
Electric potential at \(x_2=1 \mathrm{~cm}: V_2=80 \mathrm{~V}\)
The electric potential decreases uniformly.
The electric field is the negative gradient of the electric potential: \(E=-\frac{d V}{d x}\). For a uniform electric field, the electric field is constant.
\(
\begin{aligned}
& \Delta V=V_2-V_1 \\
& \Delta V=80 \mathrm{~V}-120 \mathrm{~V} \\
& \Delta V=-40 \mathrm{~V}
\end{aligned}
\)
\(
\begin{aligned}
& \Delta x=x_2-x_1 \\
& \Delta x=1 \mathrm{~cm}-(-1 \mathrm{~cm}) \\
& \Delta x=2 \mathrm{~cm}
\end{aligned}
\)
\(
\begin{aligned}
& E=-\frac{\Delta V}{\Delta x} \\
& E=-\frac{-40 \mathrm{~V}}{2 \mathrm{~cm}} \\
& E=20 \frac{\mathrm{~V}}{\mathrm{~cm}}
\end{aligned}
\)
This is the value of the electric field along the x axis.
Electric field is maximum along the direction in which the potential decreases at the maximum rate. But here, direction in which the potential decreases at the maximum rate may or may not be along the x -axis.
From the given information, the direction of maximum decrease in potential cannot be found out accurately. So, E can be greater than \(20 \mathrm{~V} / \mathrm{cm}\) in the direction of maximum decrease in potential. So, the electric field at the origin may be equal to or greater than \(20 \mathrm{Vcm}^{-1}\).

Hint

(b, d)  The quantities that do not depend on the choice of zero potential or zero potential energy are (b) Potential difference between two points and (d) Change in potential energy of a two-charge system.

Explanation:
Potential difference: Since potential difference is calculated by subtracting the potential at one point from the potential at another point, the choice of zero potential cancels out in the calculation.

Change in potential energy: Similar to potential difference, the change in potential energy depends only on the difference in potential between the initial and final states, not the absolute potential values. ©

Why the other options are incorrect:
(a) Potential at a point: The potential at a point is measured relative to an arbitrarily chosen zero potential. If you change the zero potential, the potential at that point will change as well.
(c) Potential energy of a two-charge system: The potential energy of a two-charge system depends on the individual potentials of the charges, which in turn depend on the chosen zero potential. Therefore, the potential energy of the system is affected by the choice of zero potential.

Hint

(d) The torque on the dipole due to the field may be zero.

Explanation:
Key concept: A dipole experiences a torque when placed in an electric field, but this torque is zero when the dipole’s axis is aligned with the electric field.

Important points:
The net force on a dipole can be zero if the dipole is placed in a uniform electric field, but this is not always the case with a non-uniform field generated by a point charge. ©
The orientation of the dipole relative to the electric field determines the magnitude of the torque.

Note:
As torque is given by,
\(
\tau=\vec{p} \times \vec{E}=p E \sin \theta
\)
If the dipole moment and electric field are parallel then \(\theta=0^0\). Now we have,
\(
\begin{aligned}
& \tau=p E \sin \theta \\
& \Rightarrow \tau=\mathrm{pE} \sin 0^0 \\
& \Rightarrow \tau=0
\end{aligned}
\)

Hint

(b) The magnitudes of the forces will be equal.
Explanation:
When a proton and an electron are placed in a uniform electric field, the force experienced by each particle is directly proportional to its charge. Since a proton and an electron have equal magnitudes of charge (but opposite signs), the electric force acting on them will be equal in magnitude. However, because they have different masses, their accelerations will be different. The acceleration of a particle is given by the equation \(a=F / m\), where \(F\) is the force acting on the particle and \(m\) is its mass. Since the force is the same for both proton and electron, but their masses are different, their accelerations will be different.

Why other options are incorrect:
(a) The electric forces acting on them will be equal: This statement is only partially correct. While the magnitudes of the forces are equal, the forces themselves are in opposite directions (due to the opposite charges of the proton and electron).
(c) Their accelerations will be equal: As explained above, the accelerations of the proton and electron will not be equal because of their different masses.
(d) The magnitudes of their accelerations will be equal: The accelerations are different, so their magnitudes are also different.

Hint

(c) Electric field \(E\) is proportional to the distance \(r\) from the origin: \(E \propto r\).
Electric potential at the origin is zero: \(V(0)=0\).
The relationship between electric field and electric potential is \(E=-\frac{d V}{d r}\).
Since \(E\) is proportional to \(r\), we can write \(E=k r\), where \(k\) is a constant of proportionality.
We know that \(E=-\frac{d V}{d r}\).
Substitute \(E=k r\) into the equation: \(k r=-\frac{d V}{d r}\).
Rearrange and integrate both sides:
\(d V=-k r d r\)
\(\int d V=-k \int r d r\)
\(V=-k \frac{r^2}{2}+C\)
At \(r=0, V=0\).
Substitute these values into the equation: \(0=-k \frac{0^2}{2}+C\), so \(C=0\).
\(
V=-\frac{1}{2} k r^2
\)
The electric potential is proportional to \(r^2\).

Hint

(d) When sphere A gains a positive charge, it loses electrons, and therefore its mass slightly decreases. Conversely, when sphere B gains a negative charge, it gains electrons, and its mass slightly increases.

Hint

(a) Number of electrons removed: \(n=10^{14}\)
The sphere is initially neutral.
Elementary charge: \(e=1.6 \times 10^{-19} \mathrm{C}\)
When electrons are removed, the object becomes positively charged.
The total charge is quantized and is an integer multiple of the elementary charge (\(q=+n e\)).
\(
\begin{aligned}
& q=+n e=10^{14} \times 1.6 \times 10^{-19} \\
& q=1.6 \times 10^{-5} \mathrm{C}=16 \mu \mathrm{C}
\end{aligned}
\)
Since electrons are removed, the charge is positive.

Hint

(d) A conductor has positive charge. So, there is a deficiency of electrons.
\(
\therefore \text { Number of electrons }=\frac{14.4 \times 10^{-19}}{1.6 \times 10^{-19}}=9
\)

Hint

(c) An alpha particle consists of 2 protons and 2 neutrons.
The charge of a proton is \(+1.6 \times 10^{-19} \mathrm{C}\).
The charge of a neutron is 0 C.
\(
\begin{aligned}
&\text { Charge on } \alpha \text {-particle, } q=n e\\
&\begin{aligned}
q & =+2 e=2 \times 1.6 \times 10^{-19} \\
& =3.2 \times 10^{-19} \mathrm{C}
\end{aligned}
\end{aligned}
\)

Hint

(d) According to the quantisation of charge, the charge on any body is always an integral multiple of the charge possessed by an electron. That is, if \(q\) be the charge on a body then, it can be represented as \(q=n e\), here, \(n\) is an integer and \(e\) is the charge on an electron which is equal to \(-1.6 \times 10^{-19} \mathrm{C}\).

A body is said to be positively charged when it has a shortage of electrons and negatively charges when it has an excess of electrons.

Hint

(b) When we rub glass rod with silk, excess electrons are transferred from glass to silk. So, glass rod becomes positively charged and silk becomes negatively charged.

Hint

(c) The atoms in the paper get polarized by the charged comb.

Explanation:
When a comb runs through dry hair, it generates static electricity. This charge then polarizes the molecules in the nearby paper, meaning the atoms in the paper align themselves in response to the electric field created by the comb. This polarization creates an attractive force between the charged comb and the polarized paper.

Hint

(a) When a positively charged body is earthed (connected to the earth), electrons flow from the earth into the body, meaning the body becomes uncharged.

Hint

(d) When a positive point charge is placed outside a conducting sphere, a rearrangement of charge takes place on the surface. But the total charge on the sphere is zero as no charge has left or entered the sphere.

When a positive charge \(+Q\) is placed outside a neutral conducting sphere it will induce a negative charge \(-Q\) on the side of the sphere closer to it and an equal positve charge \(+Q\) on the oppositive side of the sphere. The net charge on the sphere will be zero.

Hint

(a) Towards the plate. Here’s why:
Explanation:
When a neutral metallic sphere is placed near a positively charged plate, the electric field of the plate induces a separation of charge within the sphere. The positive charge in the sphere is repelled by the positive charge on the plate and moves away from it, while the negative charge is attracted to the plate and moves toward it. This creates a situation where the side of the sphere facing the plate becomes negatively charged, while the opposite side becomes slightly positively charged. The force of attraction between the negatively charged side of the sphere and the positive plate is stronger than the repulsion between the slightly positive side of the sphere and the positive plate, resulting in an overall attractive force towards the plate.

Key Concepts:
Induced Charge: The process of creating a separation of charge within a neutral object due to the presence of an external electric field.

Hint

(b)
\(
\begin{aligned}
&\text { According to Coulomb’s law, }\\
&F \propto \frac{1}{r^2} \Rightarrow \frac{F_1}{F_2}=\left(\frac{r_2}{r_1}\right)^2
\end{aligned}
\)
\(
\begin{aligned}
&\frac{5}{F_2}=\left(\frac{0.04}{0.06}\right)^2\\
&\text { Force between two charges, } F_2=11.25 \mathrm{~N}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& F=\frac{1}{4 \pi \varepsilon_0} \frac{\left(+7 \times 10^{-6}\right)\left(-5 \times 10^{-6}\right)}{r^2}=–\frac{1}{4 \pi \varepsilon_0} \frac{35 \times 10^{12}}{r^2} \mathrm{~N} \\
& F^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{\left(+5 \times 10^{-6}\right)\left(-7 \times 10^{-6}\right)}{r^2}=-\frac{1}{4 \pi \varepsilon_0} \frac{35 \times 10^{12}}{r^2} \mathrm{~N} \\
& F^{\prime}=F
\end{aligned}
\)

Hint

(d) \(F=K \cdot \frac{q^2}{r^2}\)
If \(q\) is halved, \(r\) is doubled, then
\(
\begin{array}{ll}
\Rightarrow & F^{\prime}=K \frac{(q / 2)^2}{(2 r)^2} \\
\Rightarrow & F^{\prime}=K \frac{q^2}{16 r^2} \\
\Rightarrow & F^{\prime}=\frac{F}{16}
\end{array}
\)
The new force acting on each charge is \(\frac{F}{16}\).

Hint

(c) Force in air: \(F_{\text {air }}=10^{-4} \mathrm{~N}\)
Force in oil: \(F_{\text {oil }}=2.5 \times 10^{-5} \mathrm{~N}\)
The force between two charges is inversely proportional to the dielectric constant of the medium.
The dielectric constant \(k\) is given by:
\(k=\frac{F_{\text {air }}}{F_{\text {oil }}}\)
Substitute the given values:
\(k=\frac{10^{-4} \mathrm{~N}}{2.5 \times 10^{-5} \mathrm{~N}}\)
Simplify:
\(k=\frac{10}{2.5}\)
\(k=4\)
The dielectric constant of oil is 4.

Hint

(c) The force \(F\) between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) in air is given by Coulomb’s law:
\(
F=\frac{q_1 q_2}{4 \pi \epsilon_0 r^2}
\)
where \(\epsilon_0\) is the permittivity of free space.
When the same charges are placed in a medium with dielectric constant \(K\), the force \(F_m\) is given by:
\(
F_m=\frac{q_1 q_2}{4 \pi \epsilon_0 K r^{\prime 2}}
\)
where \(r^{\prime}\) is the new distance between the charges in the medium.
According to the problem, we want the force in the medium to be equal to the force in air:
\(
F=F_m
\)
Substituting the expressions for \(F\) and \(F_m\) :
\(
\frac{q_1 q_2}{4 \pi \epsilon_0 r^2}=\frac{q_1 q_2}{4 \pi \epsilon_0 K r^{\prime 2}}
\)
Solving the above equation, we get
\(
r^{\prime}=\frac{r}{\sqrt{K}}
\)

Hint

(d) Distance between electrons: \(r=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Gravitational constant: \(G=6.674 \times 10^{-11} \mathrm{~N} \mathrm{~m}^2 / \mathrm{kg}^2\)
Coulomb’s constant: \(k=8.987 \times 10^9 \mathrm{~N} \mathrm{~m}^2 / \mathrm{C}^2\)
Electron mass: \(m_e=9.109 \times 10^{-31} \mathrm{~kg}\)
Electron charge: \(q_e=-1.602 \times 10^{-19} \mathrm{C}\)
Gravitational force formula: \(F_g=G \frac{m_1 m_2}{r^2}\)
Electrostatic force formula: \(F_e=k \frac{\left|q_1 q_2\right|}{r^2}\)
\(
\text { Since both particles are electrons, } m_1=m_2=m_e \text {. }
\)
Gravitational force, \(F_g=\frac{G\left(m_e\right)\left(m_e\right)}{r^2}\)
\(
\text { Since both particles are electrons, } q_1=q_2=q_e \text {. }
\)
Electrostatic force, \(F_e=\frac{1}{4 \pi \varepsilon_0} \frac{(q_e)(Q_e)}{r^2}\)
\(
\begin{aligned}
&\begin{aligned}
\therefore \quad \frac{F_g}{F_e} & =\frac{G\left(m_e\right)^2}{\left(\frac{1}{4 \pi \varepsilon_0}\right) e^2}=\frac{6.67 \times 10^{-11} \times\left(9.1 \times 10^{-31}\right)^2}{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2} \\
& =2.39 \times 10^{-43}
\end{aligned}\\
&\text { So, ratio of } F_g / F_e \text { is of order } 10^{-43} \text {. }
\end{aligned}
\)

Hint

(d) Two particles have equal mass \(m\) and charge \(q\).
The distance between them is \(r=16 \mathrm{~cm}=0.16 \mathrm{~m}\).
The net force on each particle is zero.
Gravitational force between two masses \(m_1\) and \(m_2\) separated by a distance \(r\) is \(F_g=G \frac{m_1 m_2}{r^2}\).
Electrostatic force between two charges \(q_1\) and \(q_2\) separated by a distance \(r\) is \(F_e=k \frac{q_1 q_2}{r^2}\), where \(k=\frac{1}{4 \pi \varepsilon_0}\).
Gravitational force:
\(F_g=G \frac{m^2}{r^2}, \quad m_1=m_2=m\)
Electrostatic force:
\(F_e=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2}, \quad q_1=q_2=q\)
Since the net force is zero, the magnitudes of the gravitational and electrostatic forces are equal:
\(F_g=F_e\)
\(G \frac{m^2}{r^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2}\)
Solving we get \(\frac{q}{m}=\sqrt{4 \pi \varepsilon_0 G}\)

Hint

(c) Coulomb’s law describes the electrostatic force between two point charges.
The force between two charges is independent of the presence of other charges.
The force between \(q_1\) and \(q_2\) is given by Coulomb’s law:
\(F_{12}=k \frac{\left|q_1 q_2\right|}{r^2}\)
where \(k\) is Coulomb’s constant and \(r\) is the distance between \(q_1\) and \(q_2\).
The force between \(q_1\) and \(q_2\) is still given by Coulomb’s law:
\(F_{12}^{\prime}=k \frac{\left|q_1 q_2\right|}{r^2}\)
The presence of \(q_3\) does not affect the force between \(q_1\) and \(q_2\).
The net force on \(q_2\) will change due to the force from \(q_3\), but the force exerted by \(q_1\) on \(q_2\) remains the same.
\(
\begin{aligned}
&F_{12}=F_{12}^{\prime}\\
&\text { The force exerted by } q_1 \text { on } q_2 \text { remains the same. }
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\text { Calculate the force } F_{A C} \text { between charges at } \mathrm{A} \text { and } \mathrm{C}\\
&\begin{aligned}
& F_{A C}=k \frac{\left|q_A q_C\right|}{r^2} \\
& F_{A C}=\left(9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right) \frac{\left(1 \times 10^{-6} \mathrm{C}\right)\left(2 \times 10^{-6} \mathrm{C}\right)}{(0.1 \mathrm{~m})^2} \\
& F_{A C}=1.8 \mathrm{~N}
\end{aligned}
\end{aligned}
\)
Similarly force \(F_{BC}\) \(\text { between charges at } \mathrm{B} \text { and } \mathrm{C}\)
\(
\begin{aligned}
& F_{B C}=k \frac{\left|q_B q_C\right|}{r^2} \\
& F_{B C}=\left(9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right) \frac{\left(1 \times 10^{-6} \mathrm{C}\right)\left(2 \times 10^{-6} \mathrm{C}\right)}{(0.1 \mathrm{~m})^2} \\
& F_{B C}=1.8 \mathrm{~N}
\end{aligned}
\)
Since the forces \(F_{A C}\) and \(F_{B C}\) are equal and the angle between them is \(120^{\circ}\), the resultant force is equal to the magnitude of either force.
\(
\begin{aligned}
&\text { Net force on } C \text {, }\\
&\begin{aligned}
F_{\text {net }} & =\sqrt{\left(F_{AC}\right)^2+\left(F_{BC}\right)^2+2 F_{AC} F_{BC} \cos 120^{\circ}} \\
& =\sqrt{(1.8)^2+(1.8)^2+2(1.8)(1.8)(-1 / 2)}=1.8 \mathrm{~N}
\end{aligned}
\end{aligned}
\)

Hint

(b) Given, Mass of the particle: \(m=5 \times 10^{-5} \mathrm{~kg}\)
Electric field strength: \(E=10^7 \mathrm{NC}^{-1}\)
The electric field is directed vertically downwards.
The particle is stationary.
Gravitational acceleration: \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\)
For the particle to be stationary, the net force acting on it must be zero.
The electric force on a charge \(q\) in an electric field \(E\) is given by \(F_e=q E\).
The gravitational force on a mass \(m\) is given by \(F_g=m g\).

How to solve?
Equate the magnitudes of the electric force and the gravitational force and solve for the charge \(q\).
For the particle to be stationary, the electric force must balance the gravitation force:
\(F_e=F_g\)

Since the electric field is directed downwards, for the forces to balance, the electric force must be directed upwards.
This means that the charge must be negative.
The equation becomes:
\(|q| E=m g\)
\(
q=\frac{m g}{E}=\frac{5 \times 10^{-5} \times 10}{10^7}=5 \times 10^{-5} \mu \mathrm{C}
\)
Since the electric field is directed downwards and the electric force must be upwards to counteract gravity, the charge must be negative.
Therefore, the charge is:
\(q=-5 \times 10^{-5} \mu \mathrm{C}\)

Hint

(d) Electric field due to a point charge: \(E=\frac{k Q}{r^2}\), where \(k=\frac{1}{4 \pi \epsilon_0}\).
\(
\begin{aligned}
E & =9 \times 10^9 \times \frac{Q}{r^2} \\
500 & =9 \times 10^9 \times \frac{Q}{(3)^2} \\
Q & =0.5 \mu \mathrm{C}
\end{aligned}
\)

Hint

(a)

Charge \(q_1=+5 \mu \mathrm{C}=5 \times 10^{-6} \mathrm{C}\)
Charge \(q_2=+10 \mu \mathrm{C}=10 \times 10^{-6} \mathrm{C}\)
Distance between charges \(d=20 \mathrm{~cm}=0.2 \mathrm{~m}\)
Midpoint distance from each charge \(r=\frac{d}{2}=\frac{0.2 \mathrm{~m}}{2}=0.1 \mathrm{~m}\)
Coulomb’s constant \(k=8.98755 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2} \approx 9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}\)
Electric field due to a point charge is given by \(E=k \frac{|q|}{r^2}\).
The net electric field is the vector sum of individual electric fields.
Electric field points away from positive charges.
\(
\begin{aligned}
&\text { Calculate the electric field } E_1 \text { due to charge } q_1\\
&\begin{aligned}
& E_1=k \frac{\left|q_1\right|}{r^2} \\
& E_1=\left(9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}\right) \frac{5 \times 10^{-6} \mathrm{C}}{(0.1 \mathrm{~m})^2} \\
& \left.E_1=4.5 \times 10^6 \mathrm{~N} / \mathrm{C} \text { (directed away from } q_1\right)
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
& E_2=k \frac{\left|q_2\right|}{r^2} \\
& E_2=\left(9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}\right) \frac{10 \times 10^{-6} \mathrm{C}}{(0.1 \mathrm{~m})^2} \\
& E_2=9 \times 10^6 \mathrm{~N} / \mathrm{C} \text { (directed away from } q_2 \text { ) }
\end{aligned}
\)
Since both charges are positive, the electric fields at the midpoint will be in opposite directions.
\(
\begin{aligned}
& E_{\text {net }}=\left|E_2-E_1\right| \\
& E_{\text {net }}=\left|9 \times 10^6 \mathrm{~N} / \mathrm{C}-4.5 \times 10^6 \mathrm{~N} / \mathrm{C}\right| \\
& E_{\text {net }}=4.5 \times 10^6 \mathrm{~N} / \mathrm{C}
\end{aligned}
\)
The direction of \(E_{\text {net }}\) is towards the smaller charge, \(+5 \mu \mathrm{C}\), because \(E_2>E_1\).
The net electric field at the midpoint is \(4.5 \times 10^6 \mathrm{~N} / \mathrm{C}\) directed towards the \(+5 \mu \mathrm{C}\) charge.

Hint

(c) The net field will be zero at a point outside the charges and near the charge which is smaller in magnitude.

Suppose the electric field is zero at \(P\) as shown in the figure.
Hence, at \(P, \quad \frac{k 8 q}{(L+l)^2}=\frac{k \cdot(2 q)}{l^2} \Rightarrow l=L\)
So, distance of \(P\) from origin is, \(L+L=2 L\)

Hint

(d)

A cube with side length \(b\).
A charge \(q\) at each of the 8 vertices.
The electric field due to a point charge is given by \(\vec{E}=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} \hat{r}\).
The electric field is a vector quantity.
Due to symmetry, the electric field at the center of the cube is zero.
The distance from a vertex to the center of the cube is:
\(r=\frac{\sqrt{b^2+2 b^2}}{2}=\frac{b}{2} \sqrt{3}\).

The electric field due to one charge is:
\(E_1=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}=\frac{1}{4 \pi \varepsilon_0} \frac{q}{(\sqrt{3} b)^2}=\frac{1}{4 \pi \varepsilon_0} \frac{4 q}{3 b^2}\).
For every charge at a vertex, there is an opposite charge at the opposite vertex.
The electric fields due to these opposite charges cancel each other at the center of the cube.
Since all charges are equal and symmetrically placed, the net electric field at the center is zero.

Hint

(c) Due to induction charges will appear on the surface of sphere as shown in diagram (C). Electric lines of force never intersect the any conducting surface. The electric field lines would be perpendicular to the surface of the uncharged metal sphere and slightly curved near its surface. The lines of force would also be perpendicular to the charged plates. This is because electric field lines are always perpendicular to the surface of a conductor. Additionally, the electric field inside a conductor is zero, so the lines of force cannot enter the sphere.

Hint

\(
\begin{aligned}
&\text { (a) Dipole moment, }\\
&\begin{aligned}
p=q \times 2 a & =1.6 \times 10^{-19} \times 4.3 \times 10^{-9} \\
& =6.8 \times 10^{-28} \mathrm{C}-\mathrm{m}
\end{aligned}
\end{aligned}
\)

Hint

(b) \(E_a\) : Electric field strength on the axial line of a short dipole.
\(E_e\) : Electric field strength on the equatorial line of the same short dipole.
The distance from the dipole to the point of measurement is the same for both axial and equatorial lines.
Electric field due to a dipole on the axial line: \(E_a=\frac{2 k p}{r^3}\), where \(k=\frac{1}{4 \pi \epsilon_0}, p\) is the dipole moment, and \(r\) is the distance from the center of the dipole.
Electric field due to a dipole on the equatorial line: \(E_e=\frac{k p}{r^3}\), where \(k=\frac{1}{4 \pi \epsilon_0}, p\) is the dipole moment, and \(r\) is the distance from the center of the dipole.
\(
\begin{aligned}
&\text { Divide } E_a \text { by } E_e \text { : }\\
&\frac{E_a}{E_e}=\frac{\frac{2 k p}{r^3}}{\frac{k p}{r^3}}=2
\end{aligned}
\)
\(
E_a=2 E_e
\)
The electric field strength on the axial line is twice the electric field strength on the equatorial line at the same distance.

Hint

(b) The electric field at a distance \(r\) on the axis of the dipole is given by:
\(E_0=\frac{2 k p}{r^3} \text { [along } P \text { ] }\)
The electric field at a distance \(2 r\) on the perpendicular bisector (Equatorial electric field) is given by \(\text { [opposite to } P \text { ] }\):
\(E=-\frac{k p}{(2 r)^3}\)
\(E=-\frac{k p}{8 r^3}\)
\(
\begin{aligned}
& E=-\frac{1}{8} \cdot \frac{E_0 r^3}{2 r^3} \\
& E=-\frac{E_0}{16}
\end{aligned}
\)
\(
\text { The electric field at a distance } 2 r \text { on the perpendicular bisector is }-\frac{E_0}{16}
\)

Hint

(c) The electric field at a distance \(r\) on the axial line of a dipole is:
\(E=\frac{2 k p}{r^3}\)
After rotating the dipole by \(90^{\circ}\), the point is now on the equatorial line. The electric field at a distance \(r\) on the equatorial line of a dipole is:
\(E^{\prime}=\frac{k p}{r^3}\)
Divide \(E^{\prime}[latex] by [latex]E\) :
\(\frac{E^{\prime}}{E}=\frac{\frac{k p}{r^3}}{\frac{2 k p}{r^3}}=\frac{1}{2}\)
Therefore:
\(E^{\prime}=\frac{E}{2}\)
The magnitude of the electric field at the same point after rotating the dipole by \(90^{\circ}\) is \(\frac{E}{2}\).

Hint

(a) Torque on an electric dipole in an electric field:
\(
\vec{\tau}=\vec{p} \times \vec{E}
\)
\(
\tau=p E \sin \theta \dots(1)
\)
By equation (1) it is clear that the torque will be maximum when the value of \(\sin \theta\) is maximum.
We know that the value of \(\sin \theta\) is maximum at \(\theta={9 0 ^ { \circ }}\).

Hint

(d) 

Electric dipole moment: \(\mathbf{p}\)
Electric field intensity: \(\mathbf{E}\)
Initial angle between \(\mathbf{p}\) and \(\mathbf{E}\) : \(\theta_1=90^{\circ}\)
Final angle between \(\mathbf{p}\) and \(\mathbf{E}\) : \(\theta_2=90^{\circ}+180^{\circ}=270^{\circ}\)
The work done in rotating an electric dipole in an electric field is given by \(W=-p E\left(\cos \theta_2-\cos \theta_1\right)\).
\(
\begin{aligned}
& W=-p E\left(\cos 270^{\circ}-\cos 90^{\circ}\right) \\
& W=-p E(0-0) \\
& W=0
\end{aligned}
\)
The work done in deflecting the dipole through an angle of \(180^{\circ}\) is 0.

Alternate:
\(
\begin{aligned}
\therefore \text { Work done } & =\int_{\theta_1=90^{\circ}}^{\theta_2=270^{\circ}} p E \sin \theta d \theta \\
& =[-p E \cos \theta]_{90^{\circ}}^{270^{\circ}}=0
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) Work done in rotating the dipole, }\\
&\begin{aligned}
W & =p E\left(\cos \theta_1-\cos \theta_2\right) \\
& =p E\left(\cos 0^{\circ}-\cos 180^{\circ}\right) \\
& =p E[1-(-1)]=2 p E
\end{aligned}
\end{aligned}
\)

Hint

(d) Maximum torque is given by
\(
\begin{aligned}
\tau_{\max } & =p E \quad\left[\because \sin 90^{\circ}=1\right] \\
& =(q \times d) E=(q \times 2 a) E=\left(4 \times 10^{-8} \times 2 \times 10^{-4}\right) \times 4 \times 10^8 \\
& =32 \times 10^{-4} \mathrm{~N}-\mathrm{m}
\end{aligned}
\)
Calculate the work done using the formula \(W=p E\left(\cos \theta_1-\cos \theta_2\right)\).
Assume the initial angle \(\theta_1=0^{\circ}\) and the final angle \(\theta_2=180^{\circ}\).
\(
\begin{aligned}
& W=\left(8 \times 10^{-12} \mathrm{Cm}\right)\left(4 \times 10^8 \mathrm{NC}^{-1}\right)\left(\cos 0^{\circ}-\cos 180^{\circ}\right) \\
& W=\left(32 \times 10^{-4} \mathrm{~J}\right)(1-(-1)) \\
& W=\left(32 \times 10^{-4} \mathrm{~J}\right)(2) \\
& W=64 \times 10^{-4} \mathrm{~J}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) Electric flux through the surface, }\\
&\phi=\mathbf{E} \cdot \mathbf{S}=(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}) \cdot(10 \hat{\mathbf{j}})=40 \text { unit }
\end{aligned}
\)

Hint

(d) A cube has six faces. We will denote the area of each face as \(A=a^2\), where \(a\) is the side length of the cube.
Step 3: Calculate the Electric Flux Through Each Face
The electric flux \(\Phi\) through a surface is given by:
\(
\Phi=\mathbf{E} \cdot \mathbf{A}
\)
where \(\mathbf{A}\) is the area vector of the surface.
For the face perpendicular to the \(x\)-axis (front and back faces):
Area vector for the front face: \(\mathbf{A}_{\mathbf{1}}=a^2 \hat{i}\)
Area vector for the back face: \(\mathbf{A}_{\mathbf{2}}=-a^2 \hat{i}\)
Flux through front face:
\(
\Phi_1=\mathbf{E} \cdot \mathbf{A}_1=\left(E_0 \hat{i}+E_0 \hat{j}+E_0 \hat{k}\right) \cdot\left(a^2 \hat{i}\right)=E_0 a^2
\)
Flux through back face:
\(
\begin{gathered}
\Phi_2=\mathbf{E} \cdot \mathbf{A}_2=\left(E_0 \hat{i}+E_0 \hat{j}+E_0 \hat{k}\right) \cdot\left(-a^2 \hat{i}\right)= \\
-E_0 a^2
\end{gathered}
\)
Total flux through these two faces:
\(
\Phi_{1+2}=E_0 a^2-E_0 a^2=0
\)
For the face perpendicular to the \(y\)-axis (left and right faces):
Area vector for the left face: \(\mathbf{A}_{\mathbf{3}}=-a^2 \hat{j}\)
Area vector for the right face: \(\mathbf{A}_{\mathbf{4}}=a^2 \hat{j}\)
Flux through left face:
\(
\begin{gathered}
\Phi_3=\mathbf{E} \cdot \mathbf{A}_3=\left(E_0 \hat{i}+E_0 \hat{j}+E_0 \hat{k}\right) \cdot\left(-a^2 \hat{j}\right)= \\
-E_0 a^2
\end{gathered}
\)
Flux through right face:
\(
\Phi_4=\mathbf{E} \cdot \mathbf{A}_4=\left(E_0 \hat{i}+E_0 \hat{j}+E_0 \hat{k}\right) \cdot\left(a^2 \hat{j}\right)=E_0 a^2
\)
Total flux through these two faces:
\(
\Phi_{3+4}=-E_0 a^2+E_0 a^2=0
\)
For the face perpendicular to the \(z\)-axis (top and bottom faces):
Area vector for the bottom face: \(\mathbf{A}_{\mathbf{5}}=-a^2 \hat{k}\)
Area vector for the top face: \(\mathbf{A}_{\boldsymbol{6}}=a^2 \hat{\boldsymbol{k}}\)
Flux through bottom face:
\(
\begin{gathered}
\Phi_5=\mathbf{E} \cdot \mathbf{A}_5=\left(E_0 \hat{i}+E_0 \hat{j}+E_0 \hat{k}\right) \cdot\left(-a^2 \hat{k}\right)= \\
-E_0 a^2
\end{gathered}
\)
Flux through top face:
\(
\Phi_6=\mathbf{E} \cdot \mathbf{A}_6=\left(E_0 \hat{i}+E_0 \hat{j}+E_0 \hat{k}\right) \cdot\left(a^2 \hat{k}\right)=E_0 a^2
\)
Total flux through these two faces:
\(
\Phi_{5+6}=-E_0 a^2+E_0 a^2=0
\)
Step 4: Calculate Total Electric Flux
Now, adding the flux through all six faces:
\(
\Phi_{\text {total }}=\Phi_{1+2}+\Phi_{3+4}+\Phi_{5+6}=0+0+0=0
\)
The total electric flux passing through the cube is:
\(
\Phi_{\text {total }}=0
\)

Note: Net electric flux passing from a closed surface in uniform electric field is always zero.

Hint

(b) The net electric flux is the sum of the entering and leaving fluxes:
\(\phi=\phi_2-\phi_1\)
Use Gauss’s law to relate the net flux to the enclosed charge:
\(\phi=\frac{q}{\epsilon_0}\)
\(q=\phi \epsilon_0\)
Substitute the expression for net flux into the equation:
\(q=\left(\phi_2-\phi_1\right) \epsilon_0\)
The electric charge inside the surface is \(\left(\phi_2-\phi_1\right) \epsilon_0\).

Hint

(a)
\(
\begin{aligned} \text { Flux from one face } & =\frac{1}{6}(\text { total flux }) \\
& =\frac{1}{6}\left(\frac{q}{\varepsilon_0}\right)=\frac{1}{3 \varepsilon_0} \quad[\because q=2 \mathrm{C}]
\end{aligned}
\)

Hint

(d) Inward electric flux: \(\phi_{\text {in }}=8 \times 10^3 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}\)
Outward electric flux: \(\phi_{\text {out }}=4 \times 10^3 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}\)
The formula for Gauss’s law is: \(\phi_{\text {net }}=\frac{q_{\text {enc }}}{\varepsilon_0}\)
Inward flux is considered negative, and outward flux is considered positive.
The net electric flux is the sum of the outward and inward fluxes:
\(
\begin{aligned}
& \phi_{\text {net }}=\phi_{\text {out }}+\phi_{\text {in }} \\
& \phi_{\text {net }}=4 \times 10^3 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}+\left(-8 \times 10^3 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}\right) \\
& \phi_{\text {net }}=-4 \times 10^3 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}
\end{aligned}
\)
Gauss’s law states:
\(
\phi_{\text {net }}=\frac{q_{\text {enc }}}{\varepsilon_0}
\)
Rearrange the formula to solve for the enclosed charge:
\(
q_{e n c}=\phi_{\text {net }} \varepsilon_0
\)
Substitute the net flux:
\(
\begin{aligned}
& q_{e n c}=\left(-4 \times 10^3 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1}\right) \varepsilon_0 \\
& q_{e n c}=-4 \times 10^3 \varepsilon_0 \mathrm{C}
\end{aligned}
\)
The total charge inside the surface is \(-4 \times 10^3 \varepsilon_0 \mathrm{C}\).

Hint

(b) (ii), (iii).
Explanation:
Gauss’s Law states that the electric flux through a closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space. If the flux is zero, then the enclosed charge must also be zero.
However, this doesn’t mean the electric field must be zero everywhere on the surface. It only means the total charge inside the surface is zero. The electric field can be non-zero at points on the surface if the field lines are entering the surface from one side and exiting from the other.

Why other options are incorrect:
(a) (i), (ii): While (ii) is true, (i) is not. As explained above, a zero flux does not necessarily mean a zero electric field everywhere on the surface.
(c) (ii), (iv): Option (iv) is incorrect. The condition of zero flux only tells us about the charge inside the closed surface, not the charge in the vicinity. The electric field created by charges outside the surface can contribute to the field on the surface, even if the net charge inside is zero.
(d) (i), (iii): Option (i) is incorrect. Charges can exist outside the surface and still yield zero flux.

Hint

(c) When calculating the flux of the electric field over a spherical Gaussian surface, the electric field will be due to all the charges, both inside and outside the surface. The electric field at any point on the surface is the vector sum of the fields due to all charges, regardless of where they are located. While Gauss’s law tells us that the flux only depends on the charge enclosed by the surface, the electric field itself is influenced by the total charge distribution.

Hint

\(
\begin{gathered}
\text { (d) By using Gauss’s law, } \int \mathbf{E} \cdot d \mathbf{A}=\int\left(\mathbf{E}_1+\mathbf{E}_2+\mathbf{E}_3+\mathbf{E}_4\right) \cdot d \mathbf{A} \\
=\frac{1}{\varepsilon_0}\left[Q_{\text {enclosed }}\right]=\frac{\left(q_1+q_2+q_3+q_4\right)}{\varepsilon_0}
\end{gathered}
\)

Hint

\(
\begin{aligned}
&\text { (a) Electric field, }\\
&\begin{aligned}
E & =\frac{\lambda}{2 \pi \varepsilon_0 r}=\frac{2 \lambda}{4 \pi \varepsilon_0 r} \\
\lambda & =\frac{E \times 4 \pi \varepsilon_0 r}{2}=18 \times 10^4 \times \frac{1}{9 \times 10^9} \times \frac{0.02}{2} \\
& =2 \times 10^{-7} \mathrm{C} / \mathrm{m}
\end{aligned}
\end{aligned}
\)

Hint

(a) Surface charge density, \(\sigma=\frac{q}{A}=\frac{17.7 \times 10^{-4} \mathrm{C} / \mathrm{m}^2}{200}\)
The electric field outside the sheet is given by
\(
E=\frac{\sigma}{2 \varepsilon_0}=\frac{17.7 \times 10^{-4}}{2 \times 8.85 \times 10^{-12} \times 200}=5 \times 10^5 \mathrm{~N} / \mathrm{C}
\)

Hint

(b) Electric field, \(E=\sigma / \varepsilon_0\)
\(
F=E e=\frac{\sigma e}{\varepsilon_0}=\frac{\left(2.0 \times 10^{-6}\right)\left(1.6 \times 10^{-19}\right)}{8.85 \times 10^{-12}}
\)
The work done by the electron against this force in travelling a distance \(x\) metre, \(W=F x\)
Also, \(W=KE=100 \mathrm{eV}=1.6 \times 10^{-19} \times 100 \mathrm{~J}\)
\(
\begin{aligned}
\therefore \quad \frac{\left(2.0 \times 10^{-6}\right)\left(1.6 \times 10^{-19}\right) x}{8.85 \times 10^{-12}} & =100 \times\left(1.6 \times 10^{-19}\right) \\
x & =0.44 \mathrm{~mm}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) Electric field, }\\
&\begin{aligned}
E & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{R^2}=\left(9 \times 10^9\right) \times \frac{0.2 \times 10^{-6}}{(0.25)^2} \\
& =2.88 \times 10^4 \mathrm{~N} / \mathrm{C}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) We have, } E=\sigma / \varepsilon_0\\
&\Rightarrow \quad \sigma=E \varepsilon_0=300 \times\left(8.85 \times 10^{-12}\right)=2.6 \times 10^{-9} \mathrm{C} / \mathrm{m}^2
\end{aligned}
\)

Hint

(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface.

Explanation: Gauss’s law essentially states that the total electric flux through a closed surface is directly proportional to the total charge enclosed within that surface, regardless of the charge distribution’s symmetry.

Why the other options are incorrect:
(a) Gauss’s law is valid only for symmetrical charge distributions:

While symmetrical charge distributions often make using Gauss’s law easier to calculate, the law is valid for any closed surface and charge distribution.
(b) Gauss’s law is valid only for charges placed in vacuum:

Gauss’s law is a fundamental principle of electrostatics and is independent of the presence or absence of a vacuum.
(c) The electric field calculated by Gauss’s law is the field due to the charges inside the Gaussian surface:
Gauss’s law calculates the total electric flux through a closed surface due to all charges enclosed within that surface, not just the charges inside the surface.

Hint

(d) Understanding the System:
We have an isolated metal cube, which is a conductor, and we are bringing a positive point charge \(Q\) close to it.
Effect of the Positive Charge:
When the positive charge \(Q\) is brought near the metal cube, it creates an electric field around itself. This electric field influences the charges within the metal cube.
Induction of Charges:
Due to the electric field from the positive charge \(Q[latex], the free electrons in the metal cube will be attracted towards the side of the cube that is closest to the positive charge. This results in a negative charge accumulating on that side of the cube.
Distribution of Charges:
As electrons move towards the side of the cube nearest to the positive charge, the opposite side of the cube will become positively charged due to the deficiency of electrons there. Thus, we have:
Negative charge [latex]-Q\) on the side of the cube closest to the positive charge.
Positive charge \(+Q\) on the opposite side of the cube.
Net Charge of the Cube:
Despite the separation of charges within the cube, the total charge of the cube remains unchanged. Since the cube was initially neutral (isolated), the net charge of the cube remains zero. The positive and negative charges induced on the surfaces of the cube balance each other out.
Conclusion:
Therefore, when a positive point charge \(Q\) is brought near an isolated metal cube, the cube will exhibit induced charges on its surfaces, with negative charge on the side facing the positive charge and positive charge on the opposite side, while the net charge of the cube remains zero.

Hint

(a, b, c, d) Since the non-conducting sheet M is given a uniform charge, it induces a charge in the metal rod A, which further induces a charge in the metal rod B, as shown in the figure. Hence, all the options are correct.

Hint

(b, c) The problem states that the electric flux through the closed surface is zero:
\(
\Phi=0
\)
Implication of Zero Flux:
From Gauss’s Law, if the flux is zero, then:
\(
\frac{Q_{e n c}}{\epsilon_0}=0
\)
This implies that the enclosed charge (\(Q_{enc}\)) must be zero:
\(
Q_{\text {enc }}=0
\)
The statement does not necessarily mean that the electric field ( \(E\) ) is zero everywhere on the surface. The electric field can be present, but it can also be such that the total number of electric field lines entering the surface equals the number of lines exiting, resulting in a net flux of zero. The electric field may or may not be zero everywhere on the surface; it can be non-zero but still yield zero net flux.

Hint

(a, c) The sphere encloses a dipole, i.e. two equal and opposite charges. In other words, net charge enclosed in the sphere is zero. Hence, the flux is zero through the sphere.
But the electric field at any point p on the sphere,
\(\mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{p}}{\mathrm{r}^3} \sqrt{3 \cos ^2 \theta+1}\), where \(\theta\) is the angle made by the point \(p\) with the centre of the dipole.
Hence, we can see that the field is not zero anywhere on the sphere.

Hint

(a, c) The electric flux through the hemisphere remains unchanged when the second charge \(Q\) is placed at positions \(A\) and \(C\).

Explanation:
According to Gauss’s Law, the electric flux through a closed surface is proportional to the enclosed charge. In this case, the charge \(q\) at the center is the only charge enclosed by the hemisphere. Therefore, the flux through the hemisphere is determined solely by the charge \(q\) regardless of the position of charge \(Q\).

Positions A and C :
Position A:
If charge \(Q\) is at position A , it is still on the hemisphere, so the total enclosed charge remains \(q\). The flux remains unchanged.

Position C:
If charge \(Q\) is at position C, it is also on the hemisphere, and the total enclosed charge remains \(q\). The flux remains unchanged.

Positions B and D:
Position B:
If charge \(Q\) is at position B , it is inside the hemisphere, so the total enclosed charge becomes \(q + Q\). The flux through the hemisphere would change, as more charge is enclosed.

Position D:
If charge \(Q\) is at position D, it is outside the hemisphere, and the total enclosed charge remains \(q\). The flux remains unchanged.

In conclusion, the flux through the hemisphere will remain unchanged if charge \(Q\) is placed at positions \(A\) and \(C\), as these positions do not alter the total enclosed charge within the hemisphere.

Hint

(c, d) Initially, there is no charge in the closed surface. As the wire is neutral, the flux initially is zero. Now, if we connect the battery and a current flows through it, the flux remains zero, as the number of electrons entering the surface is equal to number of electrons leaving. That is, net charge enclosed is zero and so is the flux.

Hint

(b) Placing a positive charge at point \(P\) inside a closed surface will make the flux through the surface positive. The positive & negative charges developed on different parts of conducting sphare due to induction. Hence the flux through the closed surface is positive.

Explanation:
Gauss’s Law states that the total electric flux through a closed surface is directly proportional to the net charge enclosed within that surface. If a positive charge is placed inside the surface, the number of electric field lines exiting the surface will be greater than the number entering, resulting in a positive flux.

Why other options are incorrect:
(a) will remain zero:

If there is no net charge inside the surface, the flux will be zero. However, placing a positive charge inside the surface creates a non-zero net charge.
(c) will become negative:

A negative flux occurs when more electric field lines enter the surface than leave it. Placing a positive charge inside the surface creates the opposite scenario.
(d) will become undefined:

The flux is defined as the number of electric field lines piercing a surface, and it is a definite quantity. Placing a charge inside the surface provides a definite number of field lines to count, making the flux defined.

Hint

(c) The electric field remains same for the plastic plate and the copper plate, as both are considered to be infinite plane sheets. So, it does not matter whether the plate is conducting or non-conducting.
The electric field due to both the plates,
\(
\mathrm{E}=\frac{\sigma}{\varepsilon_0}
\)

Hint

(a) Towards the plate. Here’s why:
Explanation:
When a metallic particle, even one with no net charge, is placed near a positively charged metal plate, the positive charges on the plate will repel the free electrons within the particle. This causes the electrons to move towards the end of the particle closest to the plate, creating a temporary negative charge on that side.

Key Point:
Induction:
This phenomenon is called electrostatic induction. The negative charges on the particle are induced by the presence of the positive charge on the plate.

Attraction:
The opposite charges on the particle and the plate then attract each other. The force on the particle is directed towards the plate.

Hint

(d) The electric field inside a metallic shell is zero.
Therefore, the charge \(q_1\) outside the shell does not exert any force on the charge \(q\) at the center.
The electric field inside a metallic shell is zero.
Therefore, the charge \(Q\) on the shell does not exert any force on the charge \(q\) at the center.
The net force on the charge \(q\) is the vector sum of the forces due to \(Q\) and \(q_1\). Since both forces are zero, the net force is zero.

Hint

(a) The electric flux through the sphere of radius 10 cm is given by \(\Phi_1=25 \mathrm{~V} \mathrm{~m}\).
According to Gauss’s law:
\(\Phi_1=\frac{Q_{\text {enc }}}{\epsilon_0}\)
\(Q_{e n c}=\Phi_1 \epsilon_0=25 \epsilon_0\)
The electric flux through the sphere of radius 20 cm is given by \(\Phi_2=\frac{Q_{\text {enc }}}{\epsilon_0}\).
Since the enclosed charge is the same:
\(\Phi_2=\frac{25 \epsilon_0}{\epsilon_0}\)
\(\Phi_2=25 \mathrm{Vm}\)
The electric flux over a concentric sphere of radius 20 cm is 25 V m.

Hint

(d) Initially, no charge is enclosed within the cube, so the flux is zero. As the rod enters, the enclosed charge increases linearly with time.
\(
\Phi=\frac{Q_{\text {encl }}}{\epsilon_0}=\frac{\lambda v t}{\epsilon_0} \text {, where } \lambda \text { is the linear charge density of the rod. }
\)
When the rod is fully inserted, the flux becomes constant and remains constant for the remaining L/2 length of the rod.
\(
\Phi_{\max }=\frac{\lambda \frac{L}{2}}{\epsilon_0}
\)
After that, as the rod moves out of the cube, the flux starts decreasing.
As the rod exits, the enclosed charge decreases linearly with time. The flux decreases linearly back to zero.
\(
\Phi=\frac{\lambda(L-v t)}{\epsilon_0}
\)
These processes are depicted only by curve (d).

Hint

(c) There are two surfaces surface 1 and surface 2.
\(
\begin{aligned}
\oint \vec{E} \cdot d \vec{s}= & \frac{Q_{e n c}}{\epsilon_0} \\
\int_1 \vec{E} \cdot d \vec{s}+ & \int_2 \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0} \\
& 2 \int \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0} \\
& \int \vec{E} \cdot d \vec{s}=\frac{q}{2 \epsilon_0}
\end{aligned}
\)