Sol. For macroscopic charges, e is very small, and n in very large. As q = ne, it behaves as if it were continuous.

Sol. Q will attract R.

Sol. is negative and is positive.

Sol. The electric field is vertically upward.

\

Sol. Lines of force around

$Q_{1}$ and $Q_{2}\left(Q_{1}>Q_{2}\right)$

shown in figure

Sol. At B, electric field will be minimum, as the field lines are far apart.

Sol. It is used to easily compute the electric field intensities due to certain charge systems or charged bodies, where use of coulomb’s law is quite difficult.

Sol. As, $W=p E\left(\cos \theta_{1}-\cos \theta_{2}\right)$

(i) ${{\theta }_{1}}={{0}^{{}^\circ }}\text{ }\!\!\And\!\!\text{ }{{\theta }_{2}}={{90}^{{}^\circ }},\quad W=pE$

(ii) ${{\theta }_{1}}={{0}^{{}^\circ }}\text{ }\!\!\And\!\!\text{ }{{\theta }_{2}}={{180}^{{}^\circ }},\quad W=2pE$

(iii) ${{\theta }_{1}}={{0}^{{}^\circ }}\text{ }\!\!\And\!\!\text{ }{{\theta }_{2}}={{360}^{{}^\circ }},\quad W=0$

Sol. In stable equilibrium angle between $\vec{p}$ and $\vec{E}$ is $0^{\circ}$ and in

unstable equilibrium this angle becomes $180^{\circ} .$ Work done

$=p E\left(\cos \theta_{1}-\cos \theta_{2}\right)$

Hence, $\theta_{1}=0^{\circ}, \theta_{2}=180^{\circ}=p E(1+1)=2 p E$

Sol. (i) Equidistant planes parallel to the x – y plane,

(ii) Again the equipotential surface are planes parallel to the x – y plane but they become closer together as the field increases,

(iii) Concentric spheres with centre at the origin.

Sol. As explained points A and B are at same potential. It follows that

$V_{C}-V_{A}=V_{C}-V_{B}$

Hence, work done in taking a point charge from A to C or from C to B will be the same.

Short Answer (2 Marks)

Sol. Given $P=4 \times 10^{-9} \mathrm{cm}, \theta=30^{\circ}, E=5 \times 10^{4} \mathrm{NC}^{-1}$

$\tau=P E \sin \theta=4 \times 10^{-9} \times 5 \times 10^{4} \times \sin 30^{\circ}$

$=4 \times 5 \times 10^{-5} \times \frac{1}{2}=10^{-4} \mathrm{Nm}$

Sol. Suppose the breadth of rectangular path is d. The force on the charge Q is QE. The work in taking charge from $1 \rightarrow 2$ is $Q E \times d$ and that from $3 \rightarrow 4$ will be $-Q E \times d$ No work will be done in taking the charge from $2 \rightarrow 3$ and $4 \rightarrow 1$ because along these paths the direction of the force acting on the charge is perpendicular to the displacement. Hence net work done along the rectangular path will be zero.

Sol. Let us consider an electric dipole AB and a point P on its axial line at a distance r from its centre where we have to find electric potential,

Potential at $P$ due to charge at $A, V_{P A}=-\frac{k Q}{r+a}$

Sol. When the dipole moment is in the direction of the electric field the dipole is in ‘stable’ equilibrium because as soon as it is displaced through an angle, say a torque acts upon it bringing it back along the field.

When and are in opposite directions the torque on the dipole is zero and the dipole is again in equilibrium. But, now the equilibrium is ‘unstable’ because any displacement from this position gives rise to a torque which sets the dipole in the direction of the field.

Sol. While deriving the expression for potential energy, the potential energy of the dipole is taken as zero, when it is oriented perpendicular to the direction of the electric field. In any other orientation between $\theta=0^{\circ}$ to $\theta=90^{\circ}$ the potential energy of the dipole is less than zero and hence the negative sign.

Sol. Suppose an electric dipole $A B$ of dipole moment $\vec{p}$ is kept

in uniform electric field $E$ by making angle $\theta$ with the

direction of field

Force on charge $+q$ at $B$ is $F=q E,$ along the direction of

$\vec{E}$ force on charge $-q$ at $A$ is also in opposite direction of

. Since, these two forces are parallel and equal and also acting at different points so they will form a couple which rotates the dipole in clockwise direction.

Draw a line , perpendicular distance between the forces

= $A C=A B \sin \theta=2 l \sin \theta$

As torque $\tau=$ force $\times$ perpendicular distance between forces

$\Rightarrow \tau=(q E) \times 2 l \sin \theta \Rightarrow \tau=(q \times 2 l) E \sin \theta=p E \sin \theta$

In vector form $\vec{\tau}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{E}}$

(ii) Since two equal and opposite forces are acting on dipole so net force on it is zero i.e. it will not execute translatory motion.

Sol. The electric field exerts a force QE on charge +Q and a force – QE on charge – Q of the dipole. Hence there is no net translatory force, but a torque $p E \sin \theta$ acts on the dipole which aligns the dipole parallel to the field when the dipole is released.

Sol. (a) If the charge is tripled, the flux through the surface is also tripled, because the net flux is proportional to the charge inside the surface.

(b) The flux remains constant when the volume changes. Because the surface surrounds the same amount of charge, regardless of its volume.

(c) The total flux does not change when the shape of the closed surface changes.

(d) The total flux through the closed surface remains unchanged when the charge inside the surface is moved to another location inside the surface.

Sol. The flux coming out of the surface doesn’t change because the flux through the Gaussian surface changes only when enclosed charge changes.

Sol. Suppose we have to find electric field at a distance r from the axis. Imagine a Gaussian surface (cylindrical) of radius ‘r’ and length L, in between the cylinders. Charge enclosed by the Gaussian surface is $=\lambda L$ At each point electric field vector and area vector are in the same direction. Thus,

Sol. Field due to an infinite plane sheet of charge : Consider an infinite plane sheet of charge having surface charge density $\sigma$ Let P be a point at a distance r from it where the field strength is to be determined using Gauss’ theorem. Let us consider a Gaussian surface in the form of a cylinder of length 2r and area of cross-section A arranged such that its lateral surface is parallel to the field lines and its ends are normal to the field lines as shown. For this reason.

$\int E \cdot \overrightarrow{d s}$will be zero for the lateral surface and it will be simply Eds for the end faces of the cylinder. Moreover, the charge enclosed by the Gaussian surface will be $\sigma A$.So according to Gauss’ theorem

Sol. When a vehicle moves, its body gets charged on account of friction due to air. The tyres also accumulate the charge on account of friction between the tyres and the road. The metallic ropes from the vehicle touching the ground enable the accumulated charges to flow to earth. This would otherwise be hazardous to the inflammable materials.

Sol. The special rubber tyres of air crafts are made slightly conducting so that electricity generated on account of friction between the tyres and the runway goes to earth.

Sol. Given $q_{A}=2 \mu C, q_{B}=5 \mu C, q_{C}=2 \mu C, q_{D}=-5 \mu C$

Distance of all the charges from the centre is same Charge at opposite corners are making pairs by producing electric field at the centre equal in magnitude and opposite in direction. So, net electric field at the centre, i.e., force on a charge of $1 \mu C$ will be zero.

Sol.

Sol. Force exerted on $+2 \mu C$ charge by charge at $A$

Force exerted on charge by charge at $C$$=1.35 \mathrm{N}$ along $\mathrm{A} C$

The resultant force of $\mathrm{F}_{1}$ and $\mathrm{F}_{2} \mathrm{is}=2.34 \mathrm{N}$ along $\mathrm{AM}$

Sol. (a)Suppose the equilibrium is stable ; then the test charge displaced slightly in any in any direction will experience a restoring force towards the null-point. That is, all field lines near the null-point should be directed inwards towards the null-point. That is, there is a net inward flux of electric field through a closed surface around the null-point. But by Gauss’s law, the

flux of electric field through a surface, not enclosing any charge, must be zero. Hence, the equilibrium cannot be stable.

(b) The mid-point of the line joining the two charges is a null-point. Displace a test charge from the null-point slightly along the line. There is a restoring force. But displace it say, normal to the line. You will see that the net force takes it away from the null-point. Remember, stability of equilibrium needs restoring force in all direction

Sol. This is because the atmosphere of earth is also being charged by thunder storms and lightning all over the globe approximately at the same rate of 1800 ampere. That is why atmosphere cannot become electrically neutral

Sol. (a) +q charge placed at the centre of the shell induces –q charge on the inner surface of the shell and +q of the outer surface. Thus surface charge on inner surface is

$\sigma=-\frac{q}{4 \pi r_{1}^{2}}$

(b) By Gauss’ law, the net charge on the inner surface enclosing the cavity (not having any charge) must be zero.

For a cavity of arbitrary shape, this is not enough to claim that electric field inside must be zero. The cavity may have positive and negative charges with total charge zero. In order to clear this point, we take a closed loop, part of which is inside the cavity along a field line and remaining part inside the conductor. Since, field inside the conductor is zero. A certain amount of work is done by the field in carrying a test charge over the closed loop. As this is not possible for an electrostatic field, no field lines are there inside the cavity (i.e., no field). Thus, we conclude that there will be no charge on the inner surface of the conductor, whatever be its shape.

Sol. Since the electric field is radially inward so charge on sphere will be negative i.e., Electric field at the surface of

Sol.

Sol. (a) As same charge is enclosed by the Gausian surface, electric flux remains unchanged $i e .$$-10^{3} \mathrm{Nm}^{2} / \mathrm{C}$

$\Rightarrow Q=-8.8 \times 10^{-9} C$

Sol. As we can imagine that given square is one of the faces of a cube of side 10cm.

Sol.

As electric field is conservative, work done does not depend upon path.

Sol.

Sol.   Here $Q=8 \times 10^{-3} C, Q_{0}=2 \times 10^{-9} C, r_{1}=3 \mathrm{cm}$ and

$r_{2}=4 \mathrm{cm}$

Electric potential energies at points A and B are

Hence, work done in taking the charge from point A to B

Sol.

Sol. Figure shows two charges of $2 \mu C$ and $-2 \mu C$ located at the points A and B, such that AB = 6 cm.

(i) For the given system of two charges, the equipoential surface will be a plane normal to the line AB joining the two charges and passing through its mid-point O. On any point on this plane, the potential is zero.

(ii) The electric field is in a direction from the point A to point B i.e. from the positive charge to negative charge and normal to the equipotential surface.

Sol. $Q_{1}=-1.6 \times 10^{-19} \mathrm{C}, Q_{2}=+1.6 \times 10^{-19} \mathrm{C}$

When zero of potential energy is taken at infinite separation, potential energy

Sol. In case hole is filled up, total electric field outside is

$\vec{E}_{1}+\vec{E}_{2}=\frac{\sigma}{\epsilon_{0}} \hat{n}$ …(i)

Total field inside the hollow conductor

$\vec{E}_{1}+\vec{E}_{2}=0 \cdots$ (ii)

Here $\vec{E}_{1}$ is the electric field due to the portion filling up the hole and $\vec{E}_{2}$ is the electric field due to rest of the charged conductor.

From relations (i) and (ii) we conclude that

Sol. In the I and II region of the plates, electric field is

$E_{I}=E_{I I}=0$

as electric field are equal in magnitude and opposite in directions in region I and II Electric field between the plates

Sol. As charged particles deflect towards oppositely charged plates therefore particles 1, 2 are negatively charged and particle 3 is positively charged.

Thus, particle 3 having maximum deflection along vertical direction will be having the highest charge to mass ratio.

Sol.

Sol.

Sol.

Sol. A plastic comb run through dry hairs acquires a negative charge by friction. When it is brought near a bit of paper*, the electrons in each atom of the paper are repelled away resulting in a charge separation within each atom. As the positive charge of each atom is now closer to the comb than the negative charge (which has been repelled away), the force of attraction between the comb and the positive charge of the atom exceeds the force of repulsion between the comb and the negative charge of the atom. Thus the net force on the paper is attractive.

If the hairs are wet, friction between the comb and the hairs is much reduced and so the comb does not acquire enough charge to attract bits of paper. If it is raining, the air becomes some what conducting and the comb fails to acquire charge. Hence it does not attract bits of paper.

Sol. A glass rod rubbed with silk acquires positive charge. When this rod in brought close to two uncharged metallic spheres A and B in contact negative charge in induced on left side of sphere A and an equal positive charge is induced on right side of sphere B, as shown in figure

(i) When the spheres A and B are slightly separated, the right face of A acquires positive induced charge and the

left face of B acquires negative induced charge, as shown in the figure.

ii) When the glass red rod is removed, the distribution of charges on A and B remains the same, because the positive charge on A and the negative charge on B are mutually bound.

When A and B are separated far apart, the negative and positive charges on each re-unite and both A and B become uncharged.

Sol.

Sol. Let an electric dipole AB of charge q and separation between them 2l,

The direction of electric field $\vec{E}$ isopposite to that of dipole

moment $\vec{p} .$

Electric field on axial line is double of the electric field on equitorial line for a short dipole a two equidistant points from the centre of dipole.

i.e. $E_{\text {axial }}: E_{\text {eqn }}:: 2: 1$

Sol.

Here –ve sign indicates that force is directed along –ve z-axis.

As both $\vec{P}$ and $\vec{E}$ are along $z$ -axis, in apposite direction

$\sin \theta=180^{\circ}$

$\therefore \tau=p E \sin \theta=0$

Sol.

Sol. (i) In Figure (a), as the field is due to positive charge,

$V_{P}-V_{Q}>0$

In Figure (b), as the field is due to negative charge,

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}<0$ or $\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}>0$

(ii) The potential energy of a negative charge at point Q will be negative and at point P, it will be still more negative.

Therefore,

For similar reasons, $(P . E .)_{A}-(P . E .)_{B}>0$

(iii) A small positive charge will tend to move from point P to Q and the work done by the field in moving the charge from point P to Q will be positive. Therefore, work done by electric field in moving a small positive charge from point Q to P will be negative.

(iv) For the reasons as given in (iii), work done by external agency in moving a small negative charge from B to A will be positive.

(v) As the potential energy of the negative charge increases, kinetic energy of the negative charge decreases in going from point B to A.