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**MULTIPLE CHOICE QUESTIONS - I**

**Q.1. A capacitor of 4 µ F is connected as shown in the circuit (Fig. 2.1). The internal resistance of the battery is 0.5Ω . The amount of charge on the capacitor plates will be(a) 0(b) 4µ C(c) 16µ C(d) 8µ C**

Key concept: A capacitor offers zero resistance in a circuit when it is uncharged, i.e., it can be assumed as short circuited and it offers infinite resistance when it is fully charged.

At steady state the capacitor offers infinite resistance in DC circuit and acts as open circuit as shown in figure, therefore no current flows through the capacitor and 10 Ω resistance, leaving zero potential difference across 10 Ω resistance. Hence potential difference across capacitor will be the potential difference across A and B.

The potential difference across lower and middle branch of circuit is equal to the potential difference across capacitor of upper branch of circuit.

Current flows through 2 Ω resistance from left to right, is given by I=v/R+r=1A. The potential difference across 2Ω resistance, V=IR= 1 x 2 = 2V Hence potential difference across capacitor is also 2 V.

The charge on capacitor is q = CV= (2 μF) x 2 V = 8 μC.

(a) Remains a constant because the electric field is uniform.

(b) Increases because the charge moves along the electric field.

(c) Decreases because the charge moves along the electric field.

(d) Decreases because the charge moves opposite to the electric field.

Key concept: Electric potential decreases in the direction of electric field. The direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.

The positively charged particle experiences electrostatic force along the direction of electric field, hence moves in the direction of electric field. Thus, positive work is done by the electric field on the charge. We know

Hence electrostatic potential energy of the positive charge decreases.

Key concept: For a given charge distribution, locus of all points or regions for which the electric potential has a constant value are called equipotential regions. Such equipotential can be surfaces, volumes or lines. Regarding equipotential surface the following points should be kept in mind:

- The density of the equipotential lines gives an idea about the magnitude of electric field. Higher the density, larger the field strength.
- The direction of electric field is perpendicular to the equipotential surfaces or lines.

- The equipotential surfaces produced by a point charge or a spherically charge distribution are a family of concentric spheres.
- For a uniform electric field, the equipotential surfaces are a family of plane perpendicular to the field lines.
- A metallic surface of any shape is an equipotential surface.
- Equipotential surfaces can never cross each other.
- The work done in moving a charge along an equipotential surface is always zero.

As the direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential than other equipotential surface maintained at low electrostatic potential. Hence direction of electric field is from B to A in all three cases.

The positively charged particle experiences electrostatic force along the direction of electric field, hence moves in the direction opposite to electric field. Thus, the work done by the electric field on the charge will be negative. We know

Here initial and final potentials are same in all three cases and same charge is moved, so work done is same in all three cases.**Q.4. The electrostatic potential on the surface of a charged conducting sphere is 100V. Two statments are made in this regard:S _{1} : At any point inside the sphere, electric intensity is zero.S_{2} : At any point inside the sphere, the electrostatic potential is 100V.Which of the following is a correct statement?(a) S_{1} is true but S_{2} is false.(b) Both S_{1} & S_{2} are false.(c) S_{1} is true, S_{2} is also true and S_{1} is the cause of S_{2}.(d) S_{1} is true, S_{2} is also true but the statements are independant.**

We know, the electric field intensity E and electric potential V are related

E=dV/dr

If electric field intensity E= 0, then dV/dr = 0. It means, E = 0 inside the charged conducting sphere causes uniform potential inside the sphere. Hence uniform electrostatic potential 100 V will be at any point inside the sphere.

- The electric field zero does not necessary imply that electric potential is zero. E.g., the electric field intensity at any point inside the charged spherical shell is zero but there may exist non-zero electric potential.
- If two charged particles of same magnitude but opposite sign are placed, the electric potential at the midpoint will be zero but electric field is not equal to zero.

**Q.5. Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately(a) Spheres(b) Planes(c) Paraboloids(d) Ellipsoids.**

The collection of charges, whose total sum is not zero, with regard to great distance can be considered as a single point charge. The equipotential surfaces due to a point charge are spherical.

The electric potential due to point charge q is given by V=q/4πϵ

It means electric potential due to point charge is same for all equidistant points. The locus of these equidistant points, which are at same potential, form spherical surface.

**(a) ****(b) ****(c) ****(d) ****Ans. **(c)**Solution.**

Here the system can be considered as two capacitors C_{1} and C_{2} connected in series as shown in figure.

The capacitance of parallel plate capacitor filled with dielectric block has thickness d_{1} and dielectric constant K_{2} is given by

...(i)

We can write the equivalent capacitance as

...(ii)

On comparing (i) and (ii) we have

**MULTIPLE CHOICE QUESTIONS - II**

**Q.7. Consider a uniform electric field in the direction. The potential is a constant****(a) In all space****(b) For any x for a given z****(c) For any y for a given z****(d) On the x-y plane for a given z****Ans. **(b, c, d) We know, the electric field intensity E and electric potential V are

Electric potential decreases inf the direction of electric field. The direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.

The electric field in z-direction suggest that equipotential surfaces are in x-y plane. Therefore the potential is a constant for any x for a given z, for any y for a given z and on the x-y plane for a given z.**Q.8. Equipotential surfaces(a) Are closer in regions of large electric fields compared to regions of lower electric fields.(b) Will be more crowded near sharp edges of a conductor.(c) Will be more crowded near regions of large charge densities.(d) Will always be equally spaced.**

Key concept: The density of the equipotential lines gives an idea about the magnitude of electric field. Higher the density, larger the field strength.

We know, the electric field intensity E and electric potential V are related as

or we can write

For a given

Hence the electric field intensity E is inversely proportional to the separation between equipotential surfaces. So, equipotential surfaces are closer in regions of large electric fields.)

As electric field intensities is large near sharp edges of charged conductor and near regions of large charge densities. Therefore, equipotential surfaces are closer at such places.

As the potential on equipotential surface does not change

So (V

And w = (V

And w=(V

So, work done on moving a charge is zero, verifies answer (c).

We know the work done by charge q in moving in electric field.

dW = F.dl

or answer (b) is wrong.

Answer (a) and (b) can be true only when q = +1C which is not given in question.

(a) The electric field is uniform

(b) The electric field is zero

(c) There can be no charge inside the region.

(d) The electric field shall necessarily change if a charge is placed outside the region.

We know, the electric field intensity E and electric potential V are dV related as E = - dV/dr

or we can write |E|=ΔV/Δr

The electric field intensity E and electric potential V are related as E = 0 and for V = constant, dV/dr = 0 this imply that electric field intensity E = 0.

If some charge is present inside the region then electric field cannot be zero at that region, for this V = constant is not valid.

(a) Charge on C

(b) Charge on C

(c) Charge on C

(d) Charge on C

Initially key K

By law of conservation of charge, the charge stored in capacitor Cx is equal to sum of charges on capacitors C

Q'

(a) There must be charges on the surface or inside itself

(b) There cannot be any charge in the body of the conductor

(c) There must be charges only on the surface

(d) There must be charges inside the surface

The potential of a body is due to charge of the body and due to the charge of surrounding. If tfiere are no charges anywhere else outside, then the potential of the body will be due to its own charge. If there is a cavity inside a conducting body, then charge can be placed inside the body. Hence there must be charges on its surface or inside itself. Hence option (a) is correct. The charge resides on the outer surface of a closed charged conductor. Hence there cannot be any charge in the body of the conductor. Hence option (b) is correct.

(a) In A : Q remains same but C changes.

(b) In B : V remains same but C changes.

(c) In A : V remains same and hence Q changes.

(d) In B : Q remains same and hence V changes.

The battery maintains the potential difference across connected capacitor in every circumstance. However, charge stored by disconnected charged capacitor remains conserved.

The battery maintains the potential difference across connected capacitor in every circumstance. The separation between two plates increases which in turn decreases its capacitance (C=ϵ0A/d)and potential difference across connected capacitor continue to be the same as capacitor is still connected with battery. Hence, the charge stored decreases as Q = CV.

The charge stored by isolated charged capacitor remains conserved. The separation between two plates is increasing which in turn decreases its capacitance with the decrease of capacitance, potential difference V increases as V=Q/C.

**VERY SHORT ANSWER TYPE QUESTIONS**

**Q.14. Consider two conducting spheres of radii R _{1} and R_{2} with R_{1} > R_{2}. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.**

As V

⇒

⇒

⇒

⇒

⇒

Since R

σ

i.e the charge density of smaller sphere is less than that of larger sphere

The free electrons (negative charge) experience electrostatic force in a direction opposite to the direction of electric field.

The direction of electric field is always from higher potential tolower. Hence direction of travel of electrons is from lower potential to region of higher potential.

**SHORT ANSWER TYPE QUESTIONS**

**Q.19. Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.****Ans. **Let us assume that in a closed equipotential surface with no charge the potential is changing from position to position. Let the potential just inside the surface is different to that of the surface causing in a potential gradient (dV/dr)

It means E ≠ 0 electric field comes into existence, which is given by as E=-dV/dr

It means there will be field lines pointing inwards or outwards from the surface. These lines cannot be again on the surface, as the surface is equipotential. It is possible only when the other end of the field lines are originated from the charges inside. This contradicts the original assumption. Hence, the entire volume inside must be equipotential.**Q.20. A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.****Ans. **The capacitance of the parallel plate capacitor, filled with dielectric medium of dielectric constant K is given by C=K ϵ_{0}A/d

The capacitance of the parallel plate capacitor decreases with the removal of dielectric medium as for air or vacuum K = 1 and for dielectric K > 1.

If we disconnect the battery from capacitor, then the charge stored will remain the same due to conservation of charge.

The energy stored in an isolated charge capacitor U =q^{2}/2C as q is constant, energy stored U ∝ 1/C. As C decreases with the removal of dielectric medium, therefore energy stored increases.

The potential difference across the plates of the capacitor is given by V =q/C

Since q is constant and C decreases which in turn increases V and therefore E increases as E = V/d.**Important point:****Q.21. Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity.****Ans. **The electric field E = dV/dr suggests that electric potential decreases along the direction of electric field.

Let us take any path from the eharged’conductor to the uncharged conductor along the direction of electric field. Therefore, the electric potential decrease along this path.

Now, another path from the uncharged conductor to infinity will again continually lower the potential further. This ensures that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.**Q.22. Calculate potential energy of a point charge –q placed along the axis due to a charge +Q uniformly distributed along a ring of radius R. Sketch P.E. as a function of axial distance z from the centre of the ring. Looking at graph, can you see what would happen if -q is displaced slightly from the centre of the ring (along the axis)?****Ans. **The potential energy (U) of a point charge q placed at-potential V,U=qV In our case a negative charged particle is placed at the axis of a ring having charge Q. Let the ring has radius a,the electric potential at an axial distance z from the centre of the ring is

Hence potential energy of a point charge -q is

The variation of potential energy with z is shown in the figure.

The charge -q displaced would perform oscillations. Nothing can be concluded just by looking at the graph.**Q.23. Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.****Ans. **Let us consider a ring of radius R having charge +Q distributed uniformly. Also a point P at distance z on its axis passing through centre O and perpendicular to plane of ring.

Again consider an element of ring at S of length dl having charge dq and SP is equal to r. Then potential energy due to element to r. Then potential energy due to element dl at P,

Charge on 2 Charge on 2πR length of ring =Q

Charge on dl length of ring =

So potential due to element dl at P,

dV =

Integrating over a ring the potential at P, V_{p}

**LONG ANSWER TYPE QUESTIONS**

**Q.24. Find the equation of the equipotentials for an infinite cylinder of radius r _{0}, carrying charge of linear density λ.**

The electric field lines are radial and perpendicular to the surface.

Let electric field intensity on Gaussian surface at P is E, and total charge q on cylinder will be q =

So, by Gauss’s law,

E, 2πrl cos [∠θ is between E

We know that electric field Er at distance r from centre of cylinder

So potential difference d at distance r

So equipotential surfaces are the coaxial curved surfaces of cylinders with given cylinder of radius r related as above.

Then equipotential surface will pass through S and perpendicular to line joining two chargers or AB.

So net potential at P=O

2dx = 0

2d ≠ 0

∴ x = 0

So equipotential surface will be perpendicular to X-axis passing through x=0 i.e., origin in Y-Z plane.

C

q

= α UCU

= α CU

Initial charge q

q

By the law of conservation of charge

q

78C = CU + α .CU

78 = U + α U

∴ = U+2U

Or 2U

as U is positive

Volts

Final potential on both the capacitors becomes 6 volts.

A thin conducting disc D of radius r << R of thickness ‘t’ is placed concentrically on lower plate B as shown in figure.

Let plate A and B charged with potential V.

The magnitude of electric field E between plates of capacitor

Consider Gaussian surface along circular disc D.

By Gauss’s law,

q’ is the charge conducted by plate B to disc D during charging. Nature of charge on plate B and disc will be same so repulsive force acts between B and D.

So, the charge on disc D=q’=

Electrostatic repulsive force acting on disc in upward direction

This repulsive force will be balanced by weight mg of disc D.

(b) Repeat above exercise for a proton which is made of two up and one down quark.

[charge on up quark]

Potential energy

[Talking sign of charge]

[nature sign of charges taken already]

U = -0.48 MeV

So, charges inside neutron [1q

Energy released by a neutron when converted into energy is 939 MeV.

∴ Required ratio = 0.0005111 = 5.11×10

(b) P.E. Of proton consists of 2 up and 1 down quark

r= 10

Both charged spheres are kept in contact, so charge flows between them and their potential becomes equal, let the charges on them now become q

So, V

So

Where q

By law of conservation of charges

q

5q

Hence,

∴ V

∵

Or

3V

V

From Eqns. I and II

V

3 V

V

V

∴ q

=18 x 1 μF=18 μC

q

=3 x 1 μ F x 6=18 μ C

So, charges on each capacitor i.e., q

When k

∴ q

As C

∴ q

18 μ C=3 x 1 μ F x V+3 x 1 μ F x V

18=6V

V=3Volt.

So potential on C

Now consider a ring of radius r of thickness dr on disc of radius R, as shown in figure, Again let the charge on the ring is dq then potential dV due to ring at P, will be

dq is the charge on the ring = σ .area of ring

Because dr is small therefore, dr

So the potential due to charged disc

Componendo and dividend

Then componendo and dividend of

This is the equation of sphere with centre (a,b,c) as required point is on z axis so a=0 , b=0 and z = 2d

Let equilibrium of +q is at P at a distance x from mid-point of line joining two charges.

Force F

F

∴ (d-x)

d-x = d+x (Taking square root)

-2x=0

x=0

So, Equilibrium position of charge +q between two -q charges is at mid-point (O) of line joining the two charges (-q) and (-q).

Now we have to find out potential energy of +q as a function of small distance x from balance condition (O) towards any of (-q) charge.

Let new position of charge (+q) from a small distance x from (O)

So, U is the P.E. as a function of x.

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