Electrostatic Potential and Capacitance Class 12 Important Questions: CBSE Physics Chapter 2

Very Short Answer Question (1 Mark Questions)

1. Why does the electric field inside a dielectric decrease when it is placed in an external electric field?

Ans: The electric field that is present inside a dielectric decreases when it is placed in an external electric field because of polarisation as it creates an internal electric field which is opposite to the external electric field inside a dielectric due to which the net electric field gets reduced.

2. What is the work done in moving a $2\mu C$ point change from corner A to corner B of a square ABCD when a $10\mu C$ charge exists at the centre of the square?

Ans: Point A & B are at the same distance from point O.

Hence,

$$V_A=V_B$$

$\Rightarrow \text{ Work done }=\text{0}$

3. Force of attraction between two point electric charges placed at a distance $$d$$ in a medium is $$F$$. What distance apart should these be kept in the same medium, so that force between them becomes$$\frac{F}{3}$$?

Ans: If two point charges are $$q_1$$ and $$q_2$$ separated by distance $$d$$, it can be expressed as:

$$F=\frac{K{q}_1{q}_2}{d^2}$$

Suppose if force becomes $$\frac{F}{3}$$ let the distance be $$x$$

$$\frac{F}{3}=\frac{K{q}_1{q}_2}{x^2}$$

$$\Rightarrow \frac{K{q}_1{q}_2}{3d^2}=\frac{K{q}_1{q}_2}{x^2}$$

$$\Rightarrow x^2=3d^2$$

$$\Rightarrow x=\sqrt{3}d$$

4. The distance of the field point on the equatorial plane of a small electric dipole is halved. By what factors will the electric field due to the dipole changes?

Ans: The formula $$E\propto \frac{1}{r^3}$$ gives the relation between electric field and distance. 

$$\therefore E\propto \frac{1}{{\left(\frac{r}{2}\right)}^3}\Rightarrow E\propto \frac{8}{r^3}$$

Therefore, the electric field becomes eight times larger.

5. The Plates of a charged capacitor are connected by a voltmeter. If the plates of the capacitor are moved further apart, what will be the effect on the reading of the voltmeter?

Ans: The relation between capacitance, area, distance and dielectric constant is $$C=\frac{A{\in }_0}{d}\Rightarrow C\propto \frac{1}{d}$$

Hence, if distance increases, capacitance decreases.

Since $$V=\frac{Q}{C}$$ and charge on the capacitor is constant.

Hence reading of the voltmeter increases.

6. What happens to the capacitance of a capacitor when a dielectric slab is placed between its plates?

Ans: The introduction of dielectric in a capacitor will reduce the effective charge on plate and therefore will increase the capacitance.

Short Answer Questions (2 Marks Questions)

1. Show mathematically that the potential at a point on the equatorial line of an electric dipole is Zero?

Ans:  Electric potential at point on the equatorial line of an electric dipole can be expressed mathematically as:

$$V=V_{PA}+V_{PB}$$

$$\Rightarrow V=\frac{K(-q)}{r}+\frac{K(+q)}{r}$$

$$\Rightarrow V=0$$

2. A parallel plate capacitor with air between the plates has a capacitance of $$8pF$$$$(1pF={10}^{-12}F)$$. What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant $$6$$?

Ans: For air, capacitance can be expressed as, $$C_0=\frac{A{\in }_0}{d}$$

$$C_0=8\times pF=8\times {10}^{-12}F$$

Now $$d^{\prime }=\frac{d}{2}$$ and $$K=6$$

$$\Rightarrow C^{\prime }=\frac{A{C}_0}{d^{\prime }}\times 2\times K$$

$$\Rightarrow C^{\prime }=8\times {10}^{-12}\times 2\times 6$$

$$\Rightarrow C^{\prime }=96\times {10}^{-12}pF$$

3. Draw one equipotential surfaces (1) Due to uniform electric field (2) For a point charge $$(q<0)$$?

Ans: The diagrams are given as:

4. If the amount of electric flux entering and leaving a closed surface are $${\varphi }_1$$ and $${\varphi }_2$$ respectively. What is the electric charge inside the surface?

Ans: Net flux can be given as $$={\varphi }_2-{\varphi }_1$$

Since $$\varphi =\frac{q}{{\in }_0}$$

$$\Rightarrow Q=({\varphi }_2-{\varphi }_1){\in }_0$$

The electric charge inside the surface can be given as:

$$\Rightarrow Q=\varphi {\in }_0$$

5. A stream of electrons travelling with speed $$vm/s$$ at right angles to a uniform electric field E is deflected in a circular path of radius $$r$$. Prove that $$\frac{e}{m}=\frac{v^2}{rE}$$?

Ans: The path of the electron that is travelling with velocity $$vm/s$$ at right angle of $$\bar{E}$$ is of circular shape.

It requires a centripetal in nature, $$F=\frac{m{v}^2}{r}$$

It is provided by an electrostatic force $$F=eE$$

$$\Rightarrow eE=\frac{m{v}^2}{r}$$

$$\Rightarrow \frac{e}{m}=\frac{v^2}{rE}$$

6. The distance between the plates of a parallel plate capacitor is $$d$$. A metal plate of thickness $$\left(\frac{d}{2}\right)$$ is placed between the plates. What will be the effect on the capacitance?

Ans: For air $$C_0=\frac{A{\in }_0}{d}$$

Thickness $$t=\frac{d}{2}$$ only when $$k=\mathrm{\infty }$$

$$C_0=\frac{A{\in }_0}{d}$$

$$C_{metal}=\frac{A{\in }_0}{(d-t)}$$

$$\Rightarrow C_{metal}=\frac{A{\in }_0}{(d-\frac{d}{2})}$$

$$\Rightarrow C_{metal}=2A{\in }_0$$

$$\Rightarrow C_{metal}=2C_0$$

Hence, capacitance will double.

7. Two charges $$2\mu C$$ and $$-2\mu C$$ are placed at points A and B $$6cm$$ apart.

(a) Identify an equipotential surface of the system.

Ans: The situation is represented in the adjoining figure.

An equipotential surface is the plane on which the total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.

(b) What is the direction of the electric field at every point on this surface?

Ans: The direction of the electric field at every point on this surface Is normal to the plane in the direction of AB.

8. In a Van de Graaff type generator a spherical metal shell is to be a 

$$15\times {10}^{6}V$$electrode. The dielectric strength of the gas surrounding the electrode is $$5\times {10}^{7}V{m}^{-1}$$. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)

Ans: Given that,

Potential difference is given as, $$V=15\times {10}^{6}V$$

Dielectric strength of the surrounding gas$$=5\times {10}^{7}V{m}^{-1}$$

Electric field intensity is given as, E = Dielectric strength = $$5\times {10}^{7}V{m}^{-1}$$

Minimum radius of the spherical shell required for the purpose is given by,

$$r=\frac{V}{E}$$

$$\Rightarrow r=\frac{15\times {10}^6}{5\times {10}^7}$$

$$\Rightarrow r=0.3m=30cm$$

Therefore, the minimum radius of the spherical shell required is $$30cm$$.

9. A small sphere of radius $$r_1$$ and charge $$q_1$$ is enclosed by a spherical shell of radius $$r_2$$ and charge $$q_2$$. Show that if $$q_1$$ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge $$q_2$$ on the shell is.)

Ans: According to the statement of Gauss’s law, the electric field between a sphere and a shell is determined by the charge $$q_1$$ on a small sphere. Therefore, the potential difference, $$V$$, between the sphere and the shell is independent of charge $$q_2$$. For positive charge $$q_1$$, potential difference $$V$$is always positive.

10. Describe schematically the equipotential surfaces corresponding to

(a) a constant electric field in the z-direction,

Ans: Equidistant planes which are parallel to the x-y plane are the equipotential surfaces.

(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,

Ans: Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.

(c) a single positive charge at the origin, and

Ans: Concentric spheres centred at the origin are equipotential surfaces.

(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Ans: A periodically changing shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.

Short Answer Questions (3 Marks Questions)

1. Two dielectric slabs of dielectric constant $$K_1$$ and $$K_2$$ are filled in between the two plates, each of area $$A$$, of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor? Area of each plate $$\frac{A}{2}$$.

Ans: In the question, the two capacitors are in parallel

Net Capacitance, $$C=C_1+C_2$$ 

$$C_1=\frac{K_1{\in }_0\left(\frac{A}{2}\right)}{d}=\frac{K_1{\in }_{0}A}{2d}$$

$$C_2=\frac{K_2{\in }_0\left(\frac{A}{2}\right)}{d}=\frac{K_2{\in }_{0}A}{2d}$$

$$\Rightarrow C=\frac{K_1{\in }_{0}A}{2d}+\frac{K_2{\in }_{0}A}{2d}$$

$$\Rightarrow C=\frac{{\in }_{0}A}{2d}(K_1+K_2)$$, which is the required capacitance.

2. Prove that the energy stored in a parallel plate capacitor is given by $$\frac{1}{2}C{V}^2$$.

Ans: Let us suppose a capacitor is connected to a battery and it supplies a small amount of change $$dq$$ at constant potential $$V$$, then small amount of work done by the battery is given by

$$dw=Vdq$$

$$\Rightarrow dw=\frac{qc}{dq}$$ …… (Since, $$q=CV$$)

Total work done where the capacitor is fully changed to $$q$$.

$$\int dw=W={\int }_0^q\frac{q}{c}dq$$

$$\Rightarrow W={\int }_0^{q}qdq$$

$$\Rightarrow W=\frac{q^2}{2C}=\frac{C^2{V}^2}{2C}$$

$$\Rightarrow W=\frac{1}{2}C{V}^2$$

This work done is stored in the capacitor in the form of electrostatic potential energy.

$$W=U=\frac{1}{2}C{V}^2$$

Hence proved.

3. State Gauss’s Theorem in electrostatics? Using this theorem, define an expression for the field intensity due to an infinite plane sheet of change of charge density $$\sigma C/m^2$$.

Ans: Gauss’s Theorem has a statement that electric flux through a closed surface enclosing a charge $$q$$ in vacuum is $$\frac{1}{{\in }_0}$$ times the magnitude of the charge enclosed is $$\varphi =\frac{q}{{\in }_0}$$

Let us consider a charge distributed over an infinite sheet of area S having surface change density$$\sigma C/m^2$$

To enclose the charge on sheet an imaginary Gaussian surface cylindrical in shape can be assumed and it is divided into three sections $$S_1$$, $$S_2$$ and $$S_3$$.

According to Gauss’s theorem we can infer,

$$\varphi =\frac{q}{{\in }_0}=\frac{\sigma S}{{\in }_0}$$ …… (1)

We know that, $$\varphi =\int E\cdot ds$$

For the given surface $$\varphi =\underset{S_1}{\int }\overrightarrow{E}\cdot \overrightarrow{ds}+\underset{S_2}{\int }\overrightarrow{E}\cdot \overrightarrow{ds}+\underset{S_3}{\int }\overrightarrow{E}\cdot \overrightarrow{ds}$$

$$\Rightarrow \varphi =\underset{S_1}{\int }E\cdot ds\cos 0^{\circ }+\underset{S_3}{\int }E\cdot ds\cos 0^{\circ }$$ …… $$(\because \theta ={90}^{\circ }for{S}_{2}so\underset{S_2}{\int }\overrightarrow{E}\cdot \overrightarrow{ds}=0)$$

$$\Rightarrow \varphi =E\underset{S_1}{\int }ds+E\underset{S_3}{\int }ds$$

$$\Rightarrow \varphi =E[{\int }_0^{S}ds+{\int }_0^{S}ds]$$

$$\Rightarrow \varphi =E\cdot 2S$$ ……(2)

From both equation (1) & (2)

$$E=\frac{\sigma }{2{\in }_0}$$

$$\Rightarrow E\cdot 2S=\frac{\sigma S}{{\in }_0}$$, which is the required field intensity.

4. Derive an expression for the total work done in rotating an electric dipole through an angle $$\theta$$ in a uniform electric field? $$E=1.5\times {10}^{-8}J$$

Ans: We know $$\tau =PE\sin \theta$$

If an electric dipole is rotated through an angle $$\theta$$ against the torque, then small amount of work done is

$$dw=\tau d\theta =PE\sin \theta d\theta$$

For rotating through an angle $$\theta$$ from $${90}^{\circ }$$

$$W={\int }_{{90}^{\circ }}^{\theta }PE\sin \theta$$

5. If $$C_1=3pF$$ and $$C_2=2pF$$, calculate the equivalent capacitance of the given network between A and B?

Ans: Since $$C_1$$, $$C_2$$  and $$C_1$$ are in series

$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$

$$\Rightarrow \frac{1}{C}=\frac{1}{3}+\frac{1}{2}+\frac{1}{3}$$

$$\Rightarrow \frac{1}{C}=\frac{2+3+2}{6}$$

$$\Rightarrow C=\frac{6}{7}pF$$

$$C_2$$ and C are in series

$$\Rightarrow C^{\prime }=\frac{2}{1}+\frac{6}{7}=\frac{14+6}{7}=\frac{20}{7}pF$$

Here 

$$C_1,C^{\prime }$$ and $$C_1$$ are in series

$$\frac{1}{C_{net}}=\frac{1}{C_1}+\frac{1}{C^{\prime }}+\frac{1}{C_1}$$

$$\Rightarrow \frac{1}{C_{net}}=\frac{1}{3}+\frac{7}{20}+\frac{1}{3}$$

$$\Rightarrow \frac{1}{C_{net}}=\frac{20+21+20}{60}=\frac{61}{60}$$

$$\Rightarrow \frac{1}{C_{net}}=\frac{61}{60}pF\Rightarrow C_{net}=\frac{60}{61}pF$$, which is the net capacitance.

6. Prove that energy stored per unit volume in a capacitor is given by $$\frac{1}{2}{\in }_0{E}^2$$, where E is the electric field of the capacitor.

Ans: We know capacitance of a parallel plate capacitor, $$C=\frac{A{\in }_0}{d}$$, electric field in between the plates 

Energy stored per unit volume$$=\frac{Energystored}{Volume}$$

Energy stored per unit volume$$=\frac{\frac{1}{2}\frac{q^2}{C}}{Ad}$$ (Volume of the capacitor = Ad)

$$=\frac{\frac{1}{2}\times \frac{{\left(\sigma A\right)}^2}{\frac{A{\in }_0}{d}}}{Ad}=\frac{\frac{1}{2}\times E^2{\in }^2{A}^{2}d}{A^2{\in }_{0}d}$$

$$\Rightarrow$$Energy stored/volume$$=\frac{1}{2}{\in }_0{E}^2$$. 

Therefore, proved.

7. Keeping the voltage of the charging source constant. What would be the percentage change in the energy stored in a parallel plate capacitor if the separation between its plates were to be decreased by $$10$$?

Ans: We have the relation, $$U=\frac{1}{2}C{V}^2$$

For parallel plate $$U=\frac{1}{2}\frac{A{\in }_0}{d}{V}^2$$

When $$d^{\prime }=d-\frac{10}{100}\times d=0.9d$$

Then $$U^{\prime }=\frac{1}{2}\frac{A{\in }_0}{0.9d}{V}^2$$

Change in energy$$=U^{\prime }-U=\frac{1}{2}\frac{A{\in }_0}{d}(\frac{1}{0.9}-1)$$

$$=U^{\prime }-U=u\left(\frac{0.1}{0.9}\right)=\frac{U}{9}$$

$$\Rightarrow \mathrm{\%}change=11.1$$

8. Two identical plane metallic surfaces A and B are kept parallel to each other in air separated by a distance of $$1.0cm$$ as shown in figure.

Surface A is given a positive potential of $$10V$$ and the outer surface of B is earthed.

(a) What is the magnitude and direction of the uniform electric field between point Y and Z? What is the work done in moving a charge of $$20\mu C$$ from point X to Y?

Ans:  Since $$E=\frac{dv}{dr}=\frac{10V}{1\times {10}^{-2}}=1000\frac{V}{M}$$

Since surface A is an equipotential surface i.e., $$\mathrm{\Delta }V=0$$

$\therefore$ Work done in moving a charge of $20\mu C$ from X to Y = Zero.

(b) Can we have non-zero electric potential in space, where electric field strength is zero?

Ans:  Since surface A is an equipotential surface i.e., $$\mathrm{\Delta }V=0$$

$\therefore$ Work done from X to Y = Zero.

$$E=\frac{-dv}{dr}$$ if $$E=0$$

$$\frac{dv}{dr}=0\Rightarrow dv=0$$ or V=constant (non-zero)

So, we can have non-zero electric potential, where the electric field is zero.

9. A regular hexagon of side $$10cm$$ has a charge $$5\mu C$$ at each of its vertices. Calculate the potential at the centre of the hexagon.

Ans: The given figure shows six equal numbers of charges $$q$$, at the vertices of a regular hexagon.

Where,

Charge, $$q=5\mu C=5\times {10}^{-5}C$$

Side of the hexagon, $$AB=BC=CD=DE=EF=FA=10cm$$

Distance of each vertex from center $$O$$, $$d=10cm$$

Electric potential at point $$O$$,

$$V=\frac{6\times q}{4\pi {\in }_{0}d}$$

Where,

$${\in }_0$$permittivity of free space

$$\frac{1}{4\pi {\in }_0}=9\times {10}^{9}N{C}^{-2}{m}^{-2}$$

$$V=\frac{6\times 9\times {10}^9\times 5\times {10}^{-5}}{0.1}$$

$$\Rightarrow V=2.7\times {10}^{6}V$$

Therefore, the potential at the centre of the hexagon is $$2.7\times {10}^{6}V$$.

10. A spherical conductor of radius $$12cm$$ has a charge of $$1.6\times {10}^{-7}C$$ distributed uniformly on its surface. What is the electric field? 

(a) Inside the sphere.

Ans: Radius of the spherical conductor is given as, $$r=12cm=0.12m$$

Charge is uniformly distributed over the conductor and can be given as, $$q=1.6\times {10}^{-7}C$$

Electric field inside a spherical conductor is zero. This is because if there is a field inside the conductor, then charges will move to neutralize it.

(b) Just outside the sphere 

Ans: Electric field E just outside the conductor is given by the relation,

$$E=\frac{q}{4\pi {\in }_0{r}^2}$$

Where,

$${\in }_0=$$permittivity of free space

$$\frac{1}{4\pi {\in }_0}=9\times {10}^{9}N{C}^{-2}{m}^{-2}$$

$$\Rightarrow E=\frac{1.6\times {10}^{-7}\times 9\times {10}^{-9}}{{\left(0.12\right)}^2}$$

$$\Rightarrow E={10}^{5}N{C}^{-1}$$

Therefore, the electric field just outside the sphere $$={10}^{5}N{C}^{-1}$$

(c) At a point $$18cm$$ from the centre of the sphere?

Distance of the point from the centre, $$d=18cm=0.18m$$

$$E=\frac{q}{4\pi {\in }_0{d}^2}$$

 $$\Rightarrow E=\frac{9\times {10}^9\times 1.6\times {10}^{-7}}{(18\times {10}^{-2})}$$

 $$\Rightarrow E=4.4\times {10}^{4}N/C$$

Therefore, the electric field at a point 18 cm from the centre of the sphere is $$4.4\times {10}^{4}N/C$$.

11. Three capacitors each of capacitance $$9pF$$ are connected in series.

(a) What is the total capacitance of the combination

Ans: Capacitance of each of the three capacitors, $$C=9pF$$

Equivalent capacitance $$C^{\prime }$$ of the combination of the capacitors is given by the relation,

$$\frac{1}{C}=\frac{1}{C^{\prime }}+\frac{1}{C^{\prime }}+\frac{1}{C^{\prime }}$$

$$\Rightarrow \frac{1}{C}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}$$

$$\Rightarrow C=3\mu F$$

Therefore, the total capacitance of the combination is $$3\mu F$$.

(b) What is the potential difference across each capacitor if the combination is connected to a $$120V$$ supply?

Ans:  Supply voltage is given as, $$V=100V$$

Potential difference $$V$$  across each capacitor is equal to one-third of the supply voltage.

$$V^{\prime }=\frac{V}{3}=\frac{120}{3}=40V$$

Therefore, the potential difference across each capacitor is $$40V$$.

12. Three capacitors of each capacitance $$2pF$$, $$3pF$$ and $$4pF$$  are connected in parallel.

(a) What is the total capacitance of the combination?

Ans: Capacitance of the given capacitors are

$$C_1=2pF$$

$$C_2=3pF$$

$$C_3=4pF$$

For the parallel combination of the capacitors, equivalent capacitor $$C^{\prime }$$ is given by the algebraic sum,

$$C^{\prime }=(2+3+4)pF$$

Therefore, the total capacitance of the combination is $$9pF$$.

(b) Determine the charge on each capacitor if the combination is connected to a $$100V$$ supply.

Ans:  Supply voltage, $$V=100V$$

The voltage through all the three capacitors is same as $$V=100V$$ 

Charge on a capacitor of capacitance $$C$$ and potential difference $$V$$ is given by the relation,

$$q=CV$$ …… (1)

For $$C=2pF$$,

Charge$$=VC=100\times 2=200pC=2\times {10}^{-10}C$$

For $$C=3pF$$, 

Charge$$=VC=100\times 3=300pC=3\times {10}^{-10}C$$

Now,

For $$C=4pF$$, 

Charge$$=VC=100\times 4=400pC=4\times {10}^{-10}C$$

13. In a parallel plate capacitor with air between the plates, each plate has an area and the distance between the plates is $$3mm$$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $$100V$$ supply, what is the charge on each plate of the capacitor?

Ans: Area of each plate of the parallel plate capacitor, $$A=6\times {10}^{-3}{m}^2$$

Distance between the plates, $$d=3mm=3\times {10}^{-3}m$$

Supply voltage, $$V=100V$$

Capacitance $$C$$ of a parallel plate capacitor is given by,

Where,

$$C=\frac{A{\in }_0}{d}$$

$${\in }_0=$$permittivity of free space

$$\Rightarrow {\in }_0=8.854\times {10}^{-12}{N}^{-1}{m}^{-2}{C}^{-2}$$

$$\Rightarrow C=\frac{8.854\times {10}^{-12}\times 6\times {10}^{-3}}{3\times {10}^{-3}}$$

$$\Rightarrow C=17.71\times {10}^{-12}F\Rightarrow C=17.71pF$$

Potential $$V$$ is related with the charge $$q$$ and capacitance $$C$$ as

$$V=\frac{q}{C}\Rightarrow q=VC\Rightarrow q=1.771\times {10}^{-9}C$$

$$\Rightarrow q=100\times 17.71\times {10}^{-12}C=17.71pF$$

Therefore, the capacitance of the capacitor is $$17.71pF$$ and charge on each plate is $$1.771\times {10}^{-9}$$.

14. Explain what would happen if in the capacitor given in Exercise 2.8, a $$3mm$$ thick mica sheet (of dielectric constant $$=6$$) were inserted between the plates,

(a) While the voltage supply remained connected.

Ans: Dielectric constant of the mica sheet is given in the question as, $$k=6$$

Initial capacitance is provided as, $$C=1.771\times {10}^{-11}$$

New capacitance is given as, $$C^{\prime }=kC=6\times 1.771\times {10}^{-11}=106pF$$

Supply voltage is given as, $$V=100V$$

New charge $$q^{\prime }=C^{\prime }V=6\times 1.7717\times {10}^{-9}=1.06\times {10}^{-8}Cs$$

Potential across the plates remains $$100V$$.

(b). After the supply was disconnected.

Ans: Dielectric constant is given in the question as, $$k=6$$

Initial capacitance, $$C=1.771\times {10}^{-11}F$$

New capacitance is given as $$C^{\prime }=kC=6\times 1.771\times {10}^{-11}F=106pF$$

If supply voltage is removed, then there will be no effect on the amount of charge in the plates.

Charge $$=1.771\times {10}^{-9}C$$

Potential across the plates is given by,

$$V=\frac{q}{C^{\prime }}$$

$$\Rightarrow V=\frac{1.771\times {10}^{-9}}{106\times {10}^{-12}}$$

$$\Rightarrow V=16.7V$$

15. A $$12pF$$ capacitor is connected to a $$50V$$ battery. How much electrostatic energy is stored in the capacitor?

Ans: Capacitor of the capacitance is given as, $$C=12pF=12\times {10}^{-12}F$$

Potential difference is provided as, $$V=50V$$

Electrostatic energy stored in the capacitor is given by the relation,

$$E_B=\frac{q_2}{4\pi {\in }_0{\left(BZ\right)}^2}$$

Now, 

$$\cos \theta =\frac{0.10}{0.18}=\frac{5}{9}=0.5556$$

$$\theta ={\cos }^{-1}0.5556={56.25}^{\circ }$$

$$\Rightarrow 2\theta ={112.5}^{\circ }$$

$$\Rightarrow \cos 2\theta =-0.38$$

Now, 

$$\overline{E_2}-\overline{E_1}=\frac{\sigma }{2{\in }_0}\hat{n}+\frac{\sigma }{2{\in }_0}\hat{n}=\frac{\sigma }{{\in }_0}\hat{n}$$

$$\Rightarrow E\left(2\pi dL\right)=\frac{\lambda L}{{\in }_0}$$

$$\Rightarrow E\left(2\pi dL\right)=\frac{q_1{q}_2}{4\pi {\in }_0{d}_1}-27.2eV$$

$$\Rightarrow E\left(2\pi dL\right)=\frac{9\times {10}^9\times {(1.6\times {10}^{-19})}^2}{1.06\times {10}^{-10}}-27.2eV$$

$$\Rightarrow E\left(2\pi dL\right)=21.73\times {10}^{-10}J-27.2eV$$

$$\Rightarrow E\left(2\pi dL\right)=13.58eV-27.2eV$$

$$\Rightarrow E\left(2\pi dL\right)=13.58eV$$

$$E=\frac{1}{2}C{V}^2$$

$$\Rightarrow E=\frac{1}{2}\times 12\times {10}^{-12}\times {\left(50\right)}^2$$

$$\Rightarrow E=1.5\times {10}^{-8}J$$

Therefore, the electrostatic energy stored in the capacitor is $$1.5\times {10}^{-8}J$$.

16. A $$600pF$$capacitor is charged by a $$200V$$ supply. It is then disconnected from the supply and is connected to another uncharged $$600pF$$ capacitor. How much electrostatic energy is lost in the process?

Ans: Given that,

Capacitor of the capacitance, $$C=600pF$$ 

Potential difference, $$V=200V$$

Electrostatic energy stored in the capacitor is given by,

$$E=\frac{1}{2}C{V}^2$$

$$\Rightarrow E=\frac{1}{2}\times (600\times {10}^{-12})\times {\left(200\right)}^2$$

$$\Rightarrow E=1.2\times {10}^{-5}J$$

If supply is disconnected from the capacitor and another capacitor of capacitance $$C=600pF$$ is connected to it, then equivalent capacitance $$C^{\prime }$$ of the combination is given by,

$$\frac{1}{C^{\prime }}=\frac{1}{C}+\frac{1}{C}$$

$$\Rightarrow \frac{1}{C^{\prime }}=\frac{1}{600}+\frac{1}{600}=\frac{2}{600}=\frac{1}{300}$$

$$\Rightarrow C^{\prime }=300pF$$

New electrostatic energy can be calculated as

$$E^{\prime }=\frac{1}{2}{C}^{\prime }\times V^2$$

$$E^{\prime }=\frac{1}{2}\times 300\times {\left(200\right)}^2$$

$$E^{\prime }=0.6\times {10}^{-5}J$$

Loss in electrostatic energy can be given as $$=E-E^{\prime }$$

$$\Rightarrow E-E^{\prime }=1.2\times {10}^{-5}-0.6\times {10}^{-5}$$

$$\Rightarrow E-E^{\prime }=0.6\times {10}^{-5}$$

$$\Rightarrow E-E^{\prime }=6\times {10}^{-6}J$$

Therefore, the electrostatic energy lost in the process is $$6\times {10}^{-6}J$$.

17. A spherical conducting shell of inner radius $$r_1$$ and outer radius $$r_2$$ has a charge

(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?

Ans: Charge placed at the centre of a shell is $$+q$$. Hence a charge of magnitude $$-q$$ will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is $$-q$$.

Surface charge density at the inner surface of the shell can be given by the relation,

$${\sigma }_1=\frac{TotalCharge}{OuterSurfaceArea}=\frac{-q}{4\pi {r_1}^2}$$

$${\sigma }_2=\frac{TotalCharge}{OuterSurfaceArea}=\frac{-q}{4\pi {r_2}^2}$$

A charge of $$+q$$ is induced on the outer surface of the shell. A charge of magnitude $$Q$$ is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is $$Q+q$$. Surface charge density at the outer surface of the shell.

(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but any irregular shape? Explain.

Ans:  Yes, the electric field intensity which is inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Now, take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop will be zero due to the field inside the conductor being zero. Therefore, the electric field is zero, whatever the shape.

18. If one of the two electrons of a $$H_2$$ molecule is removed, we get a hydrogen molecular ion $${H_2}^{+}$$. In the ground state of an $${H_2}^{+}$$, the two protons are separated by roughly $$1.5A^{\circ }$$, and the electron is roughly $$1A^{\circ }$$ from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Ans: The system of two protons and one electron is represented in the given figure:

Charge on proton 1, $$q_1=1.6\times {10}^{-19}C$$

Charge on proton 2, $$q_2=1.6\times {10}^{-19}C$$

Charge on electron, $$q_3=-1.6\times {10}^{-19}C$$

Distance between protons 1 and 2, $$d_1=1.5\times {10}^{-10}m$$

Distance between proton 1 and electron, $$d_2=1\times {10}^{-10}m$$

Distance between protons 2 and electron, $$d_3=1\times {10}^{-10}m$$

The potential energy at infinity is zero.

Potential energy of the system,

$$V=\frac{q_1{q}_2}{4\pi {\in }_0{d}_1}+\frac{q_1{q}_3}{4\pi {\in }_0{d}_3}+\frac{q_3{q}_1}{4\pi {\in }_0{d}_2}$$

Substituting $$\frac{1}{4\pi {\in }_{0}d}=9\times {10}^{9}N{m}^2{C}^{-2}$$

$$V=\frac{9\times {10}^9\times {10}^{-19}}{{10}^{-10}}[-{\left(16\right)}^2+\frac{{\left(1.6\right)}^2}{1.5}-{16}^2]$$

$\Rightarrow V=-30.7\times {10}^{-19}J$

$\Rightarrow V=-19.2eV$

Therefore, the potential energy of the system is $$-19.2eV$$.

19. Two charged conducting spheres of radii $$a$$ and $$b$$ are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions?

Ans: Let $$a$$ be the radius of a sphere $$A$$, $$Q_A$$ be the charge on the sphere, and $$C_A$$ be the capacitance of the sphere. Let $$b$$ be the radius of a sphere $$B$$, $$Q_B$$ be the charge on the sphere, and $$C_B$$ be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential $$V$$ will become equal.

Let $$E_A$$ be the electric field of sphere $$A$$ and $$E_B$$ be the electric field of sphere $$B$$. Therefore, their ratio,

$$\frac{E_A}{E_B}=\frac{Q_A}{4\pi {\in }_0{a}_2}\times \frac{b^{2}4\pi {\in }_0}{Q_B}$$

$$\Rightarrow \frac{E_A}{E_B}=\frac{Q_A}{Q_B}\times \frac{b^2}{a_2}$$

However, 

$$\frac{Q_A}{Q_B}=\frac{C_{A}V}{C_{B}V}$$

And 

$$\frac{C_A}{C_B}=\frac{a}{b}$$

$$\frac{Q_A}{Q_B}=\frac{a}{b}$$

Putting the value of (2) in (1), we obtain

$$\frac{E_A}{E_B}=\frac{a}{b}\frac{b^2}{a^2}=\frac{b}{a}$$

Therefore, the ratio of the electric field at the surface is $$\frac{b}{a}$$.

20. What is the area of the plates of a $$2F$$ parallel plate capacitor, given that the separation between the plates is $$0.5cm$$? (You will realize from your answer why ordinary capacitors are in the range $$\mu F$$ or less. However, electrolytic capacitors do have a much larger capacitance $$0.1F$$ because of very minute separation between the electrolytic capacitors.)

Ans: Capacitance of a parallel capacitor, $$V=2F$$

Distance between the two plates, $$d=0.5cm=0.5\times {10}^{-2}m$$. Capacitance of a parallel plate capacitor is given by the relation,

Where,

$${\in }_0=$$permittivity of free space$$=8.85\times {10}^{-12}{C}^2{N}^{-1}{m}^2$$

$$A=\frac{2\times 0.5\times {10}^{-2}}{8.85\times {10}^{-12}}$$

$$\Rightarrow A=1130k{m}^2$$

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of $$\mu F$$.

21. A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36).

Show that the capacitance of a spherical capacitor is given by$$C=\frac{4\pi {\in }_0{r}_1{r}_2}{r_1-r_2}$$ where $$r_1$$ and $$r_2$$ re the radii of outer and inner spheres, respectively.

Ans: Given that,

Radius of the outer shell is given in the question as$$=r_1$$

Radius of the inner shell is given as$$=r_2$$

The inner surface of the outer shell's charge can be given as $$+Q$$.

$$V=\frac{Q}{4\pi {\in }_0{r}_2}-\frac{Q}{4\pi {\in }_0{r}_1}$$

The outer surface of the inner shell has induced charge $$-Q$$. Potential difference between the two shells is given by,

Where,

$${\in }_0=$$permittivity of free space

$$V=\frac{Q}{4\pi {\in }_0}[\frac{1}{r_2}-\frac{1}{r_1}]$$

$$\Rightarrow V=\frac{Q(r_1-r_2)}{4\pi {\in }_0{r}_1{r}_2}$$

Capacitance of the given system is given by,

$$C=\frac{Charge\left(Q\right)}{PotentialDifference\left(V\right)}$$

$$\Rightarrow C=\frac{4\pi {\in }_0{r}_1{r}_2}{r_1-r_2}$$

Therefore, proved.

22. A cylindrical capacitor has two co-axial cylinders of length $$15cm$$ and radii $$1.5cm$$ and $$1.4cm$$. The outer cylinder is earthed and the inner cylinder is given a charge of $$3.5\mu C$$. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Ans: Given that,

Length of a co-axial cylinder is given as, $$l=15cm=0.15m$$

Radius of the outer cylinder is given as, $$r_1=1.5cm=0.015m$$

Radius of the inner cylinder is given as, $$r_2=1.4cm=0.014m$$

Charge on the inner cylinder, $$q=3.5\mu C=3.5\times {10}^{-6}C$$

Capacitance of a co-axial cylinder of radii $$r_1$$ and $$r_2$$ is given by the relation,

$$C=\frac{2\pi {\in }_{0}l}{{\log }_2\left(\frac{r_1}{r_2}\right)}$$

Where,

$${\in }_0=$$permittivity of free space$$=8.85\times {10}^{-12}{N}^{-1}{m}^{-2}{C}^2$$