# NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

## NCERT solutions for class 12 physics chapter 2 electrostatic potential and capacitance exercises:

Given, two charge particles

$q_1= 5*10^{-8}C$

$q_2= -3*10^{-8}C$

The separation between two charged particle $d=16cm=0.16m$

Now, let's assume the point P between two charged particles where the electric potential is zero is x meter away from $q_1$ and $( 9-x)$ meter away from  $q_2$

So,

The potential at point P :

$V_p=\frac{kq_1}{x}+\frac{kq_2}{0.16-x}=0$

$V_p=\frac{k5*10^{-8}}{x}+\frac{k(-3*10^{-8})}{0.16-x}=0$

$\frac{k5*10^{-8}}{x}=-\frac{k(-3*10^{-8})}{0.16-x}$

$5(0.16-x)=3x$

$x=0.1m=10cm$

Hence the point between two charged particles where the electric potential is zero lies 10cm away from $q_1$ and 6 cm away from $q_2$

Now, Let's assume a point  Q which is outside the line segment joining two charges and having zero electric potential .let the point Q lie r meter away from $q_2$ and  (0.16+r) meter away from $q_1$

So electric potential at point Q = 0

$\frac{kq_1}{0.16+r}+\frac{kq_2}{r}=0$

$\frac{k5*10^{-8}}{0.16+r}+\frac{k(-3*10^{-8})}{r}=0$

$5r=3(0.16+r)$

$r=0.24m=24cm$

Hence the second point where the electric potential is zero is 24cm away from $q_2$ and 40cm away from $q_1$

The electric potential at O due to one charge,

$V_1 = \frac{q}{4\pi\epsilon_0 r}$

q = 5 × 10-6 C

r = distance between charge and O = 10 cm = 0.1 m

Using the superposition principle, each charge at corners contribute in the same direction to the total electric potential at the point O.

$V = 6\times\frac{q}{4\pi\epsilon_0 r}$

$\implies V = 6\times\frac{9\times10^9 Nm^2C^{-2}\times5\times10^{-6}C}{0.1m}$

= $2.7 \times 10^6 V$

Therefore the required potential at the centre is $2.7 \times 10^6 V$

Given, 2 charges with charges  $2\mu C$ and $-2\mu C$.

An equipotential plane is a plane where the electric potential is the same at every point on the plane. Here if we see the plane which is perpendicular to line AB and passes through the midpoint of the line segment joining A and B, we see that at every point the electric potential is zero because the distance of all the points from two charged particles is same. Since the magnitude of charges is the same they cancel out the electric potential by them.

Hence required plane is plane perpendicular to line AB and passing through the midpoint of AB which is 3cm away from both charges.

The direction of the electric field in this surface is normal to the plane and in the direction of line joining A and B. Since both charges have the same magnitude and different sign, they cancel out the component of the electric field which is parallel to the surface.

Since the charge is uniformly distributed and it always remains on the surface of the conductor, the electric field inside the sphere will be zero.

Given,

Charge on the conductor $q=1.6*10^{-7}C$

The radius of a spherical conductor $R=12cm=0.12m$

Now,

the electric field outside the spherical conductor is given by:

$E=\frac{kq}{r^2}=\frac{1}{4\pi \epsilon }\frac{q}{r^2}=\frac{9*10^9*1.6*10^{-7}}{0.12^2}=10^5NC^{-1}$

Hence electric field just outside is $10^5NC^{-1}$.

Given,

charge on the conductor $q=1.6*10^{-7}C$

The radius of the spherical conductor $R=12cm=0.12m$

Now,

the electric field at point 18cm away from the centre of the spherical conductor is given by:

$E=\frac{kq}{r^2}=\frac{1}{4\pi \epsilon }\frac{q}{r^2}=\frac{9*10^9*1.6*10^{-7}}{0.18^2}=4.4*10^4NC^{-1}$

Hence electric field at the point 18cm away from the centre of the sphere is $4.4*10^4NC^{-1}$

As we know,

$C=\frac{\epsilon_r\epsilon_0 A}{d}$

where A= area of the plate

$\epsilon_0$ = permittivity of the free space

d = distance between the plates.

Now, Given

The capacitance between plates initially

$C_{initial}=8pF=\frac{\epsilon A}{d}$

Now, capacitance when the distance is reduced  half  and filled with the substance of dielectric 6

$C_{final}=\frac{6\epsilon_0 A}{d/2}=12\frac{\epsilon _0A}{d}=12*8pF=96pF$

Hence new capacitance is 96pF.

Given, 3 capacitor of 9pF connected in series,

the equivalent capacitance when connected in series is given by

$\frac{1}{C_{equivalent}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\frac{1}{C_{equivalent}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}$

$C_{equivalent}=3pF$

Hence total capacitance of the combination is 3pF

Given,

supply Voltage V = 120 V

The potential difference across each capacitor will be one-third of the total voltage

$V_c=\frac{V}{3}=\frac{120}{3}=40V$

Hence potential difference across each capacitor is 40V.

Given, 3 capacitors with $C_1=2pF,C_2=3pFandC_3=4pF$ are  connected in series,

the equivalent capacitance when connected in parallel is given by

$C_{equivalant}=C_1+C_2+C_3$

$C_{equivalant}=2+3+4=9pF$

Hence, the equivalent capacitance is 9pF.

Given, 3 capacitors connected in parallel with

$C_1=2pF$

$C_2=3pF$

$C_3=4pF$

Supply voltage $V=100V$

Since they are connected in parallel, the voltage across each capacitor is 100V.

So, charge on 2pf capacitor :

$Q_1=C_1V=2*10^{-12}*100=2*10^{-10}C$

Charge on 3pF capacitor:

$Q_2=C_2V=3*10^{-12}*100=3*10^{-10}C$

Charge on 4pF capacitor:

$Q_3=C_3V=4*10^{-12}*100=4*10^{-10}C$

Hence charges on capacitors are 2pC,3pC and 4pC respectively

Given,

Area of the capacitor plate $A$ = $6 \times 10^{-3}m^{2}$

Distance between the plates $d=3mm$

Now,

The capacitance of the parallel plate capacitor is

$C=\frac{\epsilon_0 A}{d}$

$\epsilon _0$= permittivity of free space =  $8.854*10^{-12}N^{-1}m^{-2}C^{_2}$

putting all know value we get,

$C=\frac{8.854*10^{-12}*6*10^{-3}}{3*10^{-3}}=17.71*10^{-12}F=17.71pF$

Hence capacitance of the capacitor is 17.71pF.

Now,

Charge on the plate of the capacitor :

$Q=CV=17.71*10^{-12}*100=1.771*10^{-9}C$

Hence charge on each plate of the capacitor is $1.771*10^{-9}C$.

Given,

The dielectric constant of the inserted mica sheet = 6

The thickness of the sheet = 3mm

Supply voltage $V = 100V$

Initial capacitance = $C_{initial}=1.771 *10^{-11}F$

Final capacitance = $KC_{initial}=6*1.771 *10^{-11}F=106*10^{-12}F$

Final charge on the capacitor = $Q_{final}=C_{final}V=106*10^{-12}*100=106*10^{-10}C$

Hence on inserting the sheet charge on each plate changes to $106*10^{-10}C$.

If a 3mm mica sheet is inserted between plates of the capacitor after disconnecting it from the power supply, the voltage across the capacitor be changed. Since the charge on the capacitor can not go anywhere, if we change the capacitance (which we are doing by inserting mica sheet here), the voltage across the capacitor has to be adjusted accordingly.

As obtained from question number 8 charge on each plate of the capacitor is $1.771*10^{-9}C$

$V_{final}=\frac{Q}{C_{final}}=\frac{1.771\times10^{-9}}{106\times 10^{-12}}=16.7V$

As we know,

the electrostatic energy stored in the capacitor is

$E=\frac{1}{2}CV^2$

Here,

$C= 12pF$

$V=50V$

So,

$E=\frac{1}{2}CV^2=\frac{1}{2}12*10^{-12}*50^2=1.5*10^{-8}J$

Hence energy stored in the capacitor is $1.5*10^{-8}J$

Given

$C=600pF$

$V=200V$

Energy stored :

$E=\frac{1}{2}CV^2=\frac{1}{2}*600*10^{-12}*200*200=1.2*10^{-5}J$

Now, when it is disconnected and connected from another capacitor of capacitance 600pF

New capacitance

$C'=\frac{600*600}{600+600}=300pF$

New electrostatic energy

$E'=\frac{1}{2}C'V^2=\frac{1}{2}*300*10^{-12}*200^2=0.6*10^{-5}J$

Hence loss in energy

$E-E'=1.2*10^{-5}-0.6*10^{-5}J=0.6*10^{-5}$

## NCERT solutions for class 12 physics chapter 2 electrostatic potential and capacitance additional exercises:

Given,

The initial distance between two charges

$d_{initial}=3cm$

The final distance between two charges

$d_{final}=4cm$

Hence total work is done

$W=q_2\left ( \frac{kq_1}{d_{final}}-\frac{kq_1}{d_{initial}}\right )=\frac{kq_1q_2}{4\pi\epsilon _0}\left ( \frac{1}{d_{final}}-\frac{1}{d_{initial}} \right )$

$W=9*10^9*8*10^{-3}*(-2*10^{-9})\left ( \frac{1}{0.04}-\frac{1}{0.03} \right )=1.27J$

The path of the charge does not matter, only initial and final position matters.

As we know,

the distance between vertices and the centre of the cube

$d=\frac{\sqrt{3}b}{2}$

Where b is the side of the cube.

So potential at the centre of the cube:

$P=8*\frac{kq}{d}=8*\frac{kq}{b\sqrt{3}/2}=\frac{16kq}{b\sqrt{3}}$

Hence electric potential at the centre will be

$\frac{16kq}{b\sqrt{3}}=\frac{16q}{4\pi \epsilon_0 b\sqrt{3}}=\frac{4q}{\pi \epsilon_0 b\sqrt{3}}$

The electric field will be zero at the centre due to symmetry i.e. every charge lying in the opposite vertices will cancel each other's field.

(a) at the mid-point of the line joining the two charges

As we know

outside the sphere, we can assume it like a point charge. so,

the electric potential at midpoint of the two-sphere

$V=\frac{kq_1}{d/2}+\frac{kq_2}{d/2}$

where q1 and q2 are charges and d is the distance between them

So,

$V=\frac{k1.5*10^{-6}}{0.15}+\frac{k2.5*10^{-6}}{0.15}=2.4*10^5V$

The electric field

$E=\frac{k1.5*10^{-6}}{0.15^2}-\frac{k2.5*10^{-6}}{0.15^2}=4*10^5V/m$

(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

The distance of the point from both the charges :

$d=\sqrt{0.1^2+0.15^2}=0.18m$

Hence,

Electric potential:

$V=\frac{kq_1}{d}+\frac{kq_2}{d}=\frac{k}{0.18}(1.5+2.5)*10^{-6}=2*10^5V$

Electric field due to q1

$E_1=\frac{kq_1}{d^2}=\frac{k1.5\mu C}{0.18^2m^2}=0.416*10^6V/m$

Electric field due to q2

$E_2=\frac{kq_2}{d^2}=\frac{k2.5\mu C}{0.18^2m^2}=0.69*10^6V/m$

Now,

Resultant Electric field :

$E=\sqrt{E_1^2+E_2^2+2E_1E_2cos\theta}$

Where $\theta$ is the angle between both electric field directions

Here,

$cos\frac{\theta}{2}=\frac{0.10}{0.18}=\frac{5}{9}$

$\frac{\theta}{2}=56.25$

${\theta}=2*56.25=112.5$

Hence

$E=\sqrt{(0.416*10^6)^2+(0.69*10^6)^2+2(0.416*10^6)(0.69*10^6)cos112.5}$

$E=6.6*10^5V/m$

(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?

The charge placed on the centre is q, so -q will be the charge induced in the inner shell and + q will be induced in the outer shell

So,

charge density on the inner shell

$\sigma_{inner}=\frac{-q}{4\pi r_1^2}$

charge Density on the outer shell

$\sigma_{outer}=\frac{Q+q}{4\pi r_2^2}$

(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.

Yes, the electric field inside the cavity is zero even when the shape is irregular and not the sphere. Suppose a Gaussian surface inside the cavity, now since there is no charge inside it, the electric flux through it will be zero according to the guess law. Also, all of the charges will reside on the surface of the conductor so, net charge inside is zero. hence electric field inside cavity is zero.

where nˆ is a unit vector normal to the surface at a point and σ is the surface charge density at that point.  Hence, show that just outside a conductor, the electric field is $\sigma \frac{\widehat{n}}{\epsilon _{0}}$

The electric field on one side of Surface with charge density $\sigma$

$E_1=-\frac{\sigma}{2\epsilon _0}\widehat{n}$

The electric field on another side of Surface with charge density $\sigma$

$E_2=-\frac{\sigma}{2\epsilon _0}\widehat{n}$

Now, resultant of both surfaces:

As E1 and E2 are opposite in direction. we have

$E_1-E_2=\frac{\sigma}{2\epsilon _0}-\left ( -\frac{\sigma}{2\epsilon _0} \right )\widehat{n}=\frac{\sigma}{\epsilon _0}$

There has to be a discontinuity at the sheet of the charge since both electric fields are in the opposite direction.

Now,

Since the electric field is zero inside the conductor,

the electric field just outside the conductor is

$E=\frac{\sigma}{\epsilon _0}\widehat{n}$

Let's assume a rectangular loop of length l and small width b.

Now,

Line integral along the loop :

$\oint E.dl=E_1l-E_2l=0$

This implies

$E_1cos\theta_1l-E_2cos\theta_2l=0$

From here,

$E_1cos\theta_1=E_2cos\theta_2$

Since $E_1cos\theta_1$ and $E_2cos\theta_2$ are the tangential component of the electric field, the tangential component of the electric field is continuous across the surface

The charge density of the cylinder with length l and radius r = $\lambda$

The radius of another hollow cylinder with same length = R

Now, let our gaussian surface be a cylinder with the same length and different radius r

the electric flux through Gaussian surface

$\oint E.ds=\frac{q}{\epsilon _0}$

$E.2\pi rl=\frac{\lambda l}{\epsilon _0}$

$E.=\frac{\lambda }{2\pi \epsilon _0r}$

Hence electric field ar a distance r from the axis of the cylinder is

$E=\frac{\lambda }{2\pi \epsilon _0r}$

(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at an infinite separation of the electron from proton

As we know,

the distance between electron-proton of the hydrogen atom

$d=0.53*10^{-10}m$

The potential energy of the system = potential energy at infinity - potential energy at distance d

$PE=0-\frac{ke*e}{d}=-\frac{9*10^9(1.6*10^{-19})^2}{0.53*10^{10}}=-43.7*10^{-19}J$

As we know,

$1ev=1.6*10^{-19}J$

$PE=\frac{-43.7*10^{-19}}{1.6*10^{-19}}=-27.2eV$

Hence potential energy of the system is -27.2eV.

(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

the potential energy of the system is -27.2eV. (from the previous question)

Kinetic energy is half of the potential energy in magnitude. so kinetic energy = 27/2 = 13.6eV

so,

total energy = 13.6 - 27.2 = -13.6eV

Hence the minimum work required to free the electron is 13.6eV

(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 $\dot{A}$ separation?

When potential energy  is zero at $d'$ 1.06 $\dot{A}$  away,

The potential energy of the system =potential energy at $d'$ -potential energy at d

$PE=\frac{ke*p}{d_1}-27.2= \frac{9*10^{9}*(1.6*10^{-19})^2}{1.06*10^{-10}}=-13.6eV$

Hence potential energy, in this case, would be -13.6eV

Given,

Distance between proton 1 and 2

$d_{p_1-p_2}=1.5*10^{-10}m$

Distance between proton 1 and electron

$d_{p_1-e}=1*10^{-10}m$

Distance between proton 2 and electron

$d_{p_2-e}=1*10^{-10}m$

Now,

The potential energy of the system :

$V=\frac{kp_1e}{d_{p_1-e}}+\frac{kp_2e}{d_{p_2-e}}+\frac{kp_1p_2}{d_{p_1-p_2}}$

Substituting the values, we get

$V=\frac{9*10^{9}*10^{-19}*10^{-19}}{10^{-10}}\left [ -(16)^2+\frac{(1.6)^2}{1.5} -(1.6)^2\right ]=-19.2eV$

Since both spheres are connected through the wire, their potential will be the same

Let electric field at A and B be $E_A and E_B$.

Now,

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

also

$\frac{Q_A}{Q_B}=\frac{C_aV}{C_BV}$

Also

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

Therefore,

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

Therefore the ratio of the electric field is b/a.

(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ?

1)electric potential at point (0,0,z)

distance from $q_1$

$d_1=\sqrt{0^2+0^2+(0-a-z)^2}=a+z$

distance from $q_2$

$d_2=\sqrt{0^2+0^2+(a-z)^2}=a-z$

Now,

Electric potential :

$V=\frac{kq_1}{a+z}+\frac{kq_2}{a-z}=\frac{2kqa}{z^2-a^2}$

2) Since the point,(x,y,0) lies in the normal to the axis of the dipole, the electric potential at this point is zero.

(b) Obtain the dependence of potential on the distance r of a point from the origin when  $\frac{r}{a}>>1$

Here, since distance r is much greater than half the distance between charges, the potential V at a distance r is inversely proportional to the square of the distance

$V\propto \frac{1}{r^2}$

(c) How much work is done in moving a small test charge from the point (5,0,0) to (–7,0,0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Since point (5,0,0) is equidistance from both charges, they both will cancel out each other potential and hence potential at this point is zero.

Similarly, point (–7,0,0) is also equidistance from both charges. and hence potential at this point is zero.

Since potential at both the point is zero, the work done in moving charge from one point to other is zero. Work done is independent of the path.

Here, As we can see

The electrostatic potential caused by the system of three charges at point P is given by

$V = \frac{1}{4\pi \varepsilon _0}\left [ \frac{q}{r+a}-\frac{2q}{r}+\frac{q}{r-a} \right ]$

$V = \frac{1}{4\pi \varepsilon _0}\left [ \frac{r(r-a)-2(r+a)(r-a)+r(r+a)}{r(r+a)(r-a)}\right ]=\frac{q}{4\pi \epsilon _0}\left [ \frac{2a^2}{r(r^2-a^2)} \right ]$

$V =\frac{q}{4\pi \epsilon _0}\left [ \frac{2a^2}{r^3(1-\frac{a^2}{r^2})} \right ]$

Since

$\frac{r}{a}>>1$

$V=\frac{2qa^2}{4\pi \epsilon _0r^3}$

From here we conclude that

$V\propto \frac{1}{r^3}$

Whereas we know that for a dipole,

$V\propto \frac{1}{r^2}$

And for a monopole,

$V\propto \frac{1}{r}$

Let's assume n capacitor connected in series and m number of such rows,

Now,

As given

The total voltage of the circuit = 1000V

and the total voltage a capacitor can withstand  = 400

From here the total number of the capacitor in series

$n=\frac{1000}{400}=2.5$

Since the number of capacitors can never be a fraction, we take n = 3.

Now,

Total capacitance required = $2\mu F$

Number of rows we need

$m=2*n=2*3=6$

Hence capacitors should be connected in 6 parallel rows where each row contains 3 capacitors in series.

Given,

The capacitance of the parallel plate capacitor $C=2F$

Separation between plated $d=0.5cm$

Now, As we know

$C=\frac{\epsilon _0A}{d}$

$A=\frac{Cd}{\epsilon _0}=\frac{2*5*10^{-3}}{8.85*10^{-12}}=1.13*10^9m^2$

$A=\1.13*10^3km^2=1130km^2$

Hence, to  get capacitance in farads, the area of the plate should be of the order od kilometre which is not good practice, and so that is why ordinary capacitors are of range $\mu F$

Given.

$C_1=100pF$

$C_2=200pF$

$C_3=200pF$

$C_4=100pF$

Now,

Lets first calculate the equivalent capacitance of $C_2\: and \:C_3$

$C_{23}=\frac{C_2C_3}{C_2+C_3}=\frac{200*200}{200+200}=100pF$

Now let's calculate the equivalent of $C_1\:and\:C_{23}$

$C_{1-23}=C_1+C_{23}=100+100=200pF$

Now let's calculate the equivalent of $C_{1-23} \: and \:C_4$

$C_{equivalent}=\frac{C_{1-23}*C_4}{C_{1-23}+C_4}=\frac{100*200}{100+200}=\frac{200}{3}pF$

Now,

The total charge on $C_4$ capacitors:

$Q_4=C_{equivalent}V=\frac{200}{3}*10^{-12}*300=2*10^{-8}C$

So,

$V_4=\frac{Q_4}{C_4}=\frac{2*10^{-8}}{100*10^{-12}}=200V$

The voltage across $C_1$ is given by

$V_{1}=V-V_{4}=300-200=100V$

The charge on $C_1$ is given by

$Q_1=C_1V_1=100*10^{-12}*100=10^{-8}C$

The potential difference across $C_2\:and\:C_3$ is

$V_2=V_3=50V$

Hence Charge on $C_2$

$Q_2=C_2V_2=200*10^{-12}*50=10^{-8}C$

And Charge on $C_3$:

$Q_3=C_3V_3=200*10^{-12}*50=10^{-8}C$

(a) How much electrostatic energy is stored by the capacitor?

Here

The capacitance of the parallel plate capacitor :

$C=\frac{\epsilon_0 A}{d}$

The electrostatic energy stored in the capacitor is given by :

$E=\frac{1}{2}CV^2=\frac{1}{2}\frac{\varepsilon _0A}{d}V^2 =\frac{1.885*10^{-12}90*10^{-4}*400^2}{2*2.5*10^{-3}}=2.55*10^{-6}J$

Hence, the electrostatic energy stored by the capacitor is $2.55*10^{-6}J$.

(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

The volume of the capacitor is:

$V=A*d=90*10^{-4}25*10^{-3}=2.25*10^{-4}m^3$

Now,

Energy stored in the capacitor per unit volume :

$u =\frac{E}{V}=\frac{2.55*10^{-6}}{2.55*10^{-4}}=0.113per \:m^3$

Now, the relation between u and E.

$u =\frac{E}{V}=\frac{\frac{1}{2}CV^2}{Ad}=\frac{\frac{1}{2}(\frac{\epsilon _0A}{d})V^2}{Ad}=\frac{1}{2}\epsilon _0E^2$

Here,

The charge on the capacitance Initially

$Q=CV=4*10^{-6}*200=8*10^{-4}C$

Total electrostatic energy initially

$E_{initial}=\frac{1}{2}CV^2=\frac{1}{2}4*10^{-6}*(200)^2=8*10^{-2}J$

Now, when it is disconnected and connected to another capacitor

Total new capacitance = $C_{new}=4+2=6\mu F$

Now, by conserving the charge on the capacitor:

$V_{new}C_{new}=C_{initial}V_{initial}$

$V_{new}6\mu F=4\mu F *200$

$V_{new}=\frac{400}{3}V$

Now,

New electrostatic energy :

$E_{new}=\frac{1}{2}C_{new}V_{new}^2=\frac{1}{2}*6*10^{-6}*\left ( \frac{400}{3} \right )^2=5.33*10^{-2}J$

Therefore,

Lost in electrostatic energy

$E=E_{initial}-E_{new}=0.08-0.0533=0.0267J$

Let

The surface charge density of the capacitor = $\sigma$

Area of the plate = $A$

Now,

As we know,

$Q=\sigma A\:and \: E=\frac{\sigma}{\epsilon _0}$

When the separation is increased by $\Delta x$,

work done by external force= $F\Delta x$

Now,

Increase in potential energy :

$\Delta u=u*A\Delta x$

By work-energy theorem,

$F\Delta x=u*A\Delta x$

$F=u*A=\frac{1}{2}\epsilon _0E^2A$

putting the value of $\epsilon _0$

$F=\frac{1}{2}\frac{\sigma}{E}E^2A=\frac{1}{2}\sigma AE=\frac{1}{2}QE$

origin of 1/2 lies in the fact that field is zero inside the conductor and field just outside is E, hence it is the average value of E/2 that contributes to the force.

Given

the radius of the outer shell = $r_1$

the radius of the inner shell = $r_2$

charge on Inner surface of outer shell =  $Q$

Induced charge on the outer surface of inner shell = $-Q$

Now,

The potential difference between the two shells

$V=\frac{Q}{4\pi \epsilon _0r_2}-\frac{Q}{4\pi \epsilon _0r_1}$

Now Capacitance is given by

$C=\frac{Charge(Q)}{Potential\:difference(V)}$

$C=\frac{Q}{\frac{Q(r_1-r_2)}{4\pi \epsilon _0r_1r_2}}=\frac{4\pi \epsilon _0r_1r_2}{r_1-r_2}$

Hence proved.

(a) Determine the capacitance of the capacitor.

The capacitance of the capacitor is given by:

$C=\frac{4\pi \epsilon _0\epsilon _rr_1r_2}{r_1-r_2}$

Here,

$C=\frac{32*0.12*0.13}{9*10^9*(0.13-0.12)}=5.5*10^{-9}F$

Hence Capacitance of the capacitor is $5.5*10^{-9}F$.

(b) what is the potential of the inner sphere?

Potential of the inner sphere is given by

$V=\frac{q}{C}=\frac{2.5*10^{-6}}{5.5*10^{-9}}=4.5*10^2$

Hence the potential of the inner sphere is $4.5*10^2 V$.

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

The radius of the isolated sphere $r = 4.5*10^2$

Now, Capacitance of sphere:

$C_{new}=4\pi\epsilon _0r=4\pi 8.85*10^{-12}*12*10^{-12}=1.33*10^{-11}F$

On comparing it with the concentric sphere, it is evident that it has lesser capacitance.this is due to the fact that the concentric sphere is connected to the earth.

Hence the potential difference is less and capacitance is more than the isolated sphere.

2.31 (a)

Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly given by $\frac{Q_{1}Q_{2}}{4\pi \epsilon_{0}r^{2}}$ , where r is the distance between their centres?

The charge on the sphere is not exactly a point charge, we assume it when the distance between two bodies is large. when the two charged sphere is brought closer, the charge distribution on them will no longer remain uniform. Hence it is not true that electrostatic force between them exactly given by $\frac{Q_{1}Q_{2}}{4\pi \epsilon_{0}r^{2}}$

2.31 (b)

If Coulomb’s law involved $\frac{1}{r^{3}}$ dependence (instead of $\frac{1}{r^{2}}$ ), would Gauss’s law be still true?

Since the solid angle is proportional to $\frac{1}{r^2}$ and not proportional to $\frac{1}{r^3}$,

The guess law which is equivalent of coulombs law will not hold true.

2.31 (c)

A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

when a small test charge is released at rest at a point in an electrostatic field configuration  it travels along the field line passing through that point only if the field lines are straight because electric field lines give the direction of acceleration, not the velocity

2.31 (d)

What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

The initial and final position will be the same for any orbit whether it is circular or elliptical. Hence work done will always be zero.

2.31 (e)

We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

Since the electric potential is not a vector quantity unlike the electric field, it can never be discontinuous.

There is no meaning in the capacitor with a single plate factually. but we give it meaning by assuming the second plate at infinity. Hence capacitance of a single conductor is the amount of change required to raise the potential of the conductor by one unit amount.

2.31 (g)

Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

Water has a much greater dielectric constant than mica because it posses a permanent dipole moment and has an unsymmetrical shape.

Given

Length of cylinder $l=15cm$

inner radius $a=1.4cm$

outer radius $b=1.5cm$

Charge on the inner cylinder $q=3.5\mu C$

Now as we know,

The capacitance of this system is given by

$C=\frac{2\pi \epsilon _0l}{2.303log_{10}(b/a)}$

$C=\frac{2\pi *8.854*10^{-12}*15*10^{-2}}{2.303log_{10}(1.5*10^{-2}/1.4*10^{-2})}=1.21*10^{-10}F$

Now

Since the outer cylinder is earthed the potential at the inner cylinder is equal to the potential difference between two cylinders.

So

Potential of inner cylinder:

$V=\frac{q}{C}=\frac{3.5*10^{-6}}{1.21*10^{-10}}=2.89*10^4V$

Given

Voltage rating in designing capacitor$V=1kV=1000V$

The dielectric constant of the material $K=\epsilon _r=3$

Dielectric strength of material = $10^7V/m$

Safety Condition:

$E=\frac{10}{100}*10^7=10^6V/m$

The capacitance of the plate $C=50pF$

Now, As we know,

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

$d=\frac{V}{E}=\frac{10^3}{10^6}=10^{-3}m$

Now,

$C=\frac{\varepsilon _0\varepsilon_rA }{d}$

$A=\frac{Cd}{\epsilon_0 \epsilon_r }=\frac{50*10^{-12}*10^{-3}}{8.85*10^{-12}*3}=1.98*10^{-3}m^2$

Hence the minimum required area is $1.98*10^{-3}m^2$

When the electric field is in the z-direction is constant, the potential in a direction perpendicular to z-axis remains constant. In other words, every plane parallel to the x-y plane is an equipotential plane.

The potential in a direction perpendicular to the direction of the field is always gonna be same irrespective of the magnitude of the electric field. Hence equipotential surface will be the plane, normal of which is the direction of the field.

For a single positive charge, the equipotential surface will be the sphere with centre at position of the charge which is origin in this case.

The equipotential surface near the grid is periodically varying.and after long distance it becomes parallel to the grid.

The potential difference between the inner sphere and shell;

$V=\frac{1}{4\pi \epsilon _0}\frac{q_1}{r_1}$

So,  the potential difference is independent of $q_2$. And hence whenever q1 is positive, the charge will flow from sphere to the shell

2.36 (a)

The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100$Vm^{-1}$. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)

The surface of the earth and our body, both are good conductors. So our body and the ground both have the same equipotential surface as we are connected from the ground. When we move outside the house, the equipotential surfaces in the air changes so that our body and ground is kept at the same potential. Therefore we do not get an electric shock.

2.36 (b)

(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. Will he get an electric shock if he touches the metal sheet next morning?

Yes, the man will get an electric shock. the aluminium sheet is gradually charged up by discharging current of atmosphere.eventually the voltage will increase up to a certain point depending on the capacitance of the capacitor formed by aluminium sheet, insulating slab and the ground. When the man touches the that charged metal, he will get a shock.

2.36 (c)

c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

Thunderstorm and lightning across the globe keep the atmosphere charged by releasing the light energy, heat energy, and sound energy in the atmosphere. In a way or other, the atmosphere is discharged through regions of ordinary weather. on an average, the two opposing currents are in equilibrium. Hence the atmosphere perpetually remains charged.

## Chapter list for class 12 physics is given below-

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions Solutions of NCERT class 12 physics chapter 11 Dual nature of radiation and matter CBSE NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei Solutions of NCERT class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

## Importance of NCERT solutions for class 12 physics chapter 2 electrostatic potential and capacitance in board exams:

The CBSE NCERT solutions for class 12 physics chapter 2 electrostatic potential and capacitance will help in securing a good score in board exam. On an average 4 to 6 % questions are asked from this chapter for the board exam. For JEE Main on an average 6% of questions are asked from electrostatics and for NEET exam 4 to 6% questions are asked from the unit electrostatics. The unit electrostatics contains the first two chapters of class 12 NCERT physics.

• Q.

### Will NCERT solve all the doubts pertaining to competitive exams?

NCERT can solve up to 70% of the doubts pertaining to competitive exams. Along with NCERT we can solve previous questions and use any one good reference books for more questions and concepts. For conceptual clarity NCERT is the best. Also solving NCERT questions will help in clearing the concept.

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