# Electrostatics Class 12 Board Questions with Answers-Physics

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**Q. Why can we ignore quantisation of electric charge when dealing with macroscopic charges.**[NCERT]

*n*in very large. As

*q = ne*, it behaves as if it were continuous.

**Q. Consider three charged bodies**

*P*,*Q*and*R*. If*P*and*Q*repel each other and*P*attracts*R*, what will be the nature of force between*Q*and*R*.*Q*will attract

*R*.

**Q. Consider the situation shown in the figure. What are the signs of Q1 and Q2?**

**Q.**The weight of a positively-charged particle is balanced by producing electric field between two parallel plates. State the direction of the electric field.

**Q. Sketch the electric lines of force for two positive**

**charge $Q_{1}$ and $Q_{2}\left(Q_{1}>Q_{2}\right)$ separated by a distance $d$.**

**Q. You are given electric lines of forces as shown :**

**At what point (A, B or C) is electric field minimum.**

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

**Q. What is the use of Gaussian surface ?**

**Q. An electric dipole of moment**

*p*is aligned parallel to a uniform electric field*E*. How much work will be done by an external field in rotating the dipole through**(i) 90o**

**(ii) 180o (end for end)**

**(iii) 360o from the direction of field.**

**Q. How much work is required in turning an electric dipole of dipole moment $\vec{P}$ from its position of stable equilibrium to its position of unstable equilibrium in a uniform electrostatic field $\vec{E}$ .**

**Q. What will be the equipotential surfaces corresponding to**(i) A constant electric field in the z-direction. (ii) A field that uniformly increase in magnitude but its direction remains unchanged along the

*z*-axis.

*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.

**Q.**

**If a point charge +**

*Q*, is taken first from*A*to*C*and then form*C*to*B*of a circle drawn with another point-charge +*Q*as centre, then along which path more work will be done?*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)**

**Q. An electric dipole with dipole moment $4 \times 10^{-9} \mathrm{C}^{m}$ is aligned at $30^{\circ}$ with direction of uniform electric field of magnitude $5 \times 10^{4} N C^{-1}$ . Calculate the magnitude of the torque acting on the dipole.**[NCERT]

**Q. A uniform electric field**

*E*exists between two charged plates. What will be the work done in moving a charge*Q*along a closed rectangular path?*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.

**Q. Derive an expression for the potential at a point along the axial line of a short 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 $A, V_{P A}=-\frac{k Q}{r+a}$

**Q. An electric dipole is situated in a uniform electric field such that its moment is aligned in the direction of the field . Is the equilibrium of the dipole stable or unstable ? If and are in opposite directions, then ?**

**Q. What does the negative sign in the expression for potential energy $(U=-p E \cos \theta)$ signify $?$**

**Q. An electric dipole is placed in uniform external field field :**

**(i) Show that the torque on the dipole moment of the dipole**

**(ii) Show that no translatory force act on it.**

**OR**

**Derive an expression for the maximum torque acting on an electric dipole, when held in a uniform electric field.**[NCERT]

*i.e.*it will not execute translatory motion.

**Q. An electric dipole is held at an angle $\theta$ in a uniform external electric field $\vec{E}$. Will there be any (i) net translatory force, (ii) torque acting on it ? Explain what happens to the dipole on being released.**

*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.

**Q. A spherical surface surrounds a point charge. Describe what happens to the total flux through the surface if : (a) the charge is tripled (b) the volume of the sphere is doubled (c) the surface is changed to a cube and (d) the charge is moved to some other location inside the surface.**

**Q. A spherical rubber balloon carries a charge that is uniformly distributed over its surface. As the balloon is blown it’s surface. As the balloon is blown up and increases in size, how does the total electric flux coming out of the surface change? Give reason.**

**Q. A long charged cylinder of linear charged density**[NCERT]

**is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders.**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,

**Q. Using Gauss’s theorem, derive an expression for electric field intensity at a point near a thin infinite plane sheet of electric charge.**

*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 2

*r*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

**Q. Vehicles carrying inflammable material usually have metallic chains touching the ground during motion. Why?**

**Q. Ordinary rubber is an insulator. But the special rubber tyres of air crafts are made slightly conducting. Why is this necessary ?**[NCERT]

**Q. Four point charges**[NCERT] $q_{A}=2 \mu C, q_{B}=-5 \mu C, q_{C}=2 \mu C,$ and $q_{D}=-5 \mu C$ are located at the corners of a square

*ABCD*of side 10

*cm*. What is the force on a charge of placed at the centre of the square.

*i.e.*, force on a charge of $1 \mu C$ will be zero.

**Q. A polythene piece rubbed with wool is found to have a negative charge of $3 \times 10^{-7} C$**

**(i) Estimate the number of electrons transferred (from which to which ?)**

**(ii) It there a transfer of mass from wool to polythene?**[NCERT]

**Q. There point charges of $+2 \mu C,-3 \mu C$ and $-3 \mu C$ are kept**

**at the vertices, $A, B$ and $C$ respectively of an equilateral**

**triangle of side $20 c m$ as shown in the figure.**

**What should be the sign and magnitude of the charge to be placed at the mid-point (**

*M*) of side*BC*so that the charge at*A*remains in equilibrium ?**Q. (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a point (i.e., where E = 0) of the configuration. Show that equilibrium of test charge is necessarily unstable.**

**(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.**[NCERT]

**Q. The discharging current in the atmosphere due to small conductivity of air is known to be 1800**[NCERT]

*A*on an average over the globe. Why then does the atmosphere not dischrge itself completely in due course and become electrically neutral ?**Q. A spherical conducting shell of inner radius $r_{1}$ and outer radius $r_{2}$ has a charge Q.**[NCERT] a) A charge

*q*is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell. (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

*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.

**Q. A conducting sphere of radius 10**[NCERT]

*cm*has an unknown charge. If the electric field 20*cm*from the centre of the sphere is $1.5 \times 10^{3} N / C$ and points radially inward, what is the value of the point charge ?*i.e.,*Electric field at the surface of

**Q. An infinite line charge produces a field of $9 \times 10^{4} N / C$ at a distance of 2**[NCERT]

*cm*. Calculate the linear charge density.**Q. A point charge causes an electric flux of $-1.0 \times 10^{3} N m^{2} / C$ to pass through a spherical Gausian surface of 10.0**

*cm*radius centred on the charge.**(a) If the radius of the Gausian surface were doubled, how much flux would pass through the surface ?**

**(b) What is the value of the point charge?**[NCERT]

**Q. A point charge $+10 \mu C$ is a distance $5 \mathrm{cm}$ directly above the centre of square of side $10 \mathrm{cm}$ as shown in Fig. What is the magnitude of the electric flux through the square ?**[NCERT]

*cm*.

**Q. (i) Calculate the potential at a point**

*P*due to charge of $4 \times 10^{-7} C$ located 9*cm*away.**(ii) Hence, obtain the work done in bringing a charge of $2 \times 10^{-9} C$ from infinity to the point**[NCERT]

*P*. Does the answer depend on the path along which the charge is brought?**Q. (i) Determine the electrostatic potential energy of a system containing two charges $7 \mu C$ and $-2 \mu C$separated by a distance of 18**

*cm*.**(ii) How much work is required to separate the two charges infinitely away from each other?**[NCERT]

**Q. A charge of $8 m C$ is located at the origin. Calculated the**

**work done is taking a small charge of $-20 \times 10^{-9} C$ from a**

**point $A(0,0,3 \mathrm{cm})$ to a point $B(0,4 \mathrm{cm}, 0),$ via point**

**$C(0,6 \mathrm{cm}, 9 \mathrm{cm})$.**[NCERT]

*A*and

*B*are Hence, work done in taking the charge from point

*A*to

*B*

**Q. If one of the two electrons from a $H_{2}$ molecule is removed, we get hydrogen molecular ion $\left(H_{2}^{+}\right)$. In the ground state of a , the two protons are separated by roughly , and the electron is roughly from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy**. [NCERT]

**Q. Two charges $2 \mu C$ and $-2 \mu C$ are placed at points**

*A*and*B*6*cm*apart. (i) Identify equipotential surface of the system. (ii) What is the direction of the electric field at every point on this surface ?*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.

**Q. In a hydrogen atom, the electron and proton are bound at a distance of about $0.53 A1.06 A$ separation ?**[NCERT]

**Q. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left(\frac{\sigma}{2 \epsilon_{0}}\right) \hat{n}$ where $\hat{n}$ is the unit vector in the outward normal direction and $\sigma$ is the surface charge density near the hole.**[NCERT]

**Q. 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 $17.0 \times 10^{-12} \mathrm{C} / \mathrm{m}^{2}$. What is**

*E*:**(a) in the outer region of the first plate**

**(b) in the outer region of the second plate, and**

**(c) between the plates**[NCERT]

**Q. Figure given below 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 ? Explain.**[NCERT]

**Q. An oil drop of 12 excess electrons is held stationary under a constant electric field of $2.55 \times 10^{-4} N C^{-1}$**

**oil drop experiment. The density of the oil is $1.26 g \mathrm{cm}^{-1}$. Estimate the radius of the drop . $\left(g=9.81 m s^{-2}, e=1.60 \times 10^{-19} \mathrm{C}\right)$**[NCERT]

**Q. (i) Two insulated charged copper spheres**

*A*and*B*have their centres separated by 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.**(ii) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved ?**[NCERT]

**Q. Suppose the spheres**[NCERT]

*A*and*B*in the above question have identical sizes.*A*third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between*A*and*B*?**Q. A plastic comb run through one’s dry hairs attracts small bits of paper. Why? What happens if the hairs are wet or if it is raining ?**[NCERT]

*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.

**Q. A glass rod rubbed with silk is brought close to two uncharged metallic spheres**[NCERT]

*A*and*B*touching each other. The rod induces charges on*A*and*B*, as shown. State what happens when : (i)*A*and*B*are slightly separated, (ii) the glass rod is then taken away, and finally (iii)*A*and*B*are separated far apart.*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.

**Q. Consider a uniform electric field $E=3 \times 10^{3} \hat{i} N / C$**

**(a) What is the flux of this 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 30° angle with the**[NCERT]

*x*-axis**Q. Derive an expression for the magnitude of electric field intensity at any point along the equatorial line of a short electric dipole. Give the direction of electric field intensity at that point. For a short dipole what is the ratio of electric field intensities at two equidistant points from the centre of dipole. One along the axial line and another on the equatorial line.**

*AB*of charge

*q*and separation between them 2

*l,*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. $E_{\text {axial }}: E_{\text {eqn }}:: 2: 1$

**Q. In a certain region of space, electric field is along the**[NCERT]

*z*-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive*z*-direction, at the rate of $10^{5} N C^{-1}$ per meter. What are the force and total dipole moment equal to $10^{-7} \mathrm{Cm}$ in the negative*z*-direction?*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 $\sin \theta=180^{\circ}$ $\therefore \tau=p E \sin \theta=0$

**Q. A molecule of a substance has permanent electric dipole moment equal to $10^{-29} C \times m$ . A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of the magnitude $\left(10^{6} \mathrm{Vm}^{-1}\right)$ . The direction of the field is suddenly changed by an angle of 60°. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of sample.**[NCERT]

**Q. Figures (a) and (b) show the field lines of a single positive and negative charge respectively :**

**(i) Give the sign of the potential difference $V_{P}-V_{Q}$ and $V_{B}-V_{A}$**

**(ii) Give the sign of the potential energy difference of a small negative charge between the point**

*Q*and*P*;*A*and*B*.**(iii) Give the sign of the work done by the field in moving a small positive charge from point**

*Q*to*P*.**(iv) Give the sign of the work done by an external agency in moving a small negative charge from point**

*B*to*A*.**(v) Does the kinetic energy of a small negative charge increase or decrease in going from point**

*B*to*A*?*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*.

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