**Question 1**
Calculate the amount of work done in assembling charge together to form a uniformly charged sphere.

Solution
Consider a sphere to be assembled by number of infinitesimally thin spherical shells. Suppose at any instant r be the radius of the sphere and now we add a charged shell of radius dr to this sphere of radius r. This process continues till the radius of sphere becomes equal to R. Now, charge on sphere of radius r having volume charge density $\rho$ is

$q= \frac {4}{3} \pi r^3 \rho$

And amount of charge on the shell of thickness dq is

$dq= 4 \pi r^2 \rho dr$

Work done in adding charge q’ to sphere containing charge q is

dW=Potential of sphere of radius r x charge contained in the shell. Thus,

$dW=\frac{\frac{4}{3}\pi r^3\rho}{4\pi\epsilon_0r} \times 4\pi r^2\rho dr$

or

$dW=\fra

c{4}{3}\frac{\pi\rho^2r^4dr}{\epsilon_0}$

Total work done in assembling a sphere of radius R would be

W=\int_{0}^{R}{\frac{4}{3}\frac{\pi\rho^2r^4dr}{\varepsilon_0}=\frac{4\pi\rho^2r^5}{15\varepsilon_0}}

If Q is the total amount of charge on the sphere then,

$Q=\frac{4}{3} \pi R^3 \rho$

$\rho=\frac{3Q}{4\pi R^3}$

Putting this value of $\rho$ in equation (1) we get

$W=\frac{4\pi R^5}{15\varepsilon_0}\left(\frac{3Q}{4\pi R^3}\right)^2$

This work is stored as potential energy of the system, hence

$U=\frac{3}{5}\frac{Q^2}{4\pi\varepsilon_0R}$

Thus, energy required to assemble a sphere of charge is directly proportional to the square of total charge and inversely proportional to the radius of the sphere

**Question 2**
A charge Q is distributed over two concentric hollow spheres of radius r and R(>r) such that the surface densities of both the spheres are equal. Find the potential at the ir common centre.

Solution
Suppose the charges on the sphere of radius r and R are $Q_1$ and $Q_2$ respectively. Then total charge,

$Q=Q_1+Q_2$

Let $\sigma$ be the surface charge density then,

$\sigma=\frac{Q_1}{4\pi r^2}=\frac{Q_2}{4\pi R^2}$

or

$\frac{Q_1}{Q_2}=\frac{r^2}{R^2}$

$\frac{Q_1}{Q_2}+1=\frac{r^2}{R^2}+1$

$\frac{Q_1+Q_2}{Q_2}=\frac{Q}{Q_2}=\frac{r^2+R^2}{R^2}$

$Q_2=Q\left(\frac{R^2}{r^2+R^2}\right)$

Similarly we can find

$Q_1=Q\left(\frac{r^2}{r^2+R^2}\right)$

Suppose at common centre, potentials due to charge $Q_1$ and $Q_2$ are $V_1$ and $V_2$. Then,

$V_1=\frac{Q_1}{4\pi\varepsilon_0r}=\frac{Qr}{4\pi\varepsilon_0r(r^2+R^2)}$

and

$V_2=\frac{Q_2}{4\pi\varepsilon_0R}=\frac{QR}{4\pi\varepsilon_0R(r^2+R^2)}$

$\therefore V=V_1+V_2=\frac{Q}{4\pi\varepsilon_0}\left(\frac{R+r}{r^2+R^2}\right)$

**Question 3**
The potential in the region of space near the point P (-2,4,6) is

$V=80x^2 + 60y^2$ V

(a) Find out the electric Field vector in the region

(b) Find out the Electric field vector at point P

(c) What is the value of potential at point P

Solution
(a) We have

$E_x=-\frac{\partial V}{\partial x} =-160x$

$E_y=-\frac{\partial V}{\partial y}=-120y$

$E_z=-\frac{\partial V}{\partial z}=0$

So Electric Field vector

=(-160x)**i**+(-120y) **j**

(b)Electric Field vector at point P==320**i**-480**j**

(c) Value of potential at point P is

$V=80x^2 + 60y^2 = 320+960 = 1280$ V

**Question 4**
Two circular wire loops of radii .09 m (loop I) and .05 m(Loop II) are placed such that their axes coincide and their center are .12 m apart. Charge of 10

^{-6} C is distributed uniformly on each loop. Find the potential difference between the centers of loop

Solution
Potential at the centre of circular loop is given by

$= (\frac{1}{4\pi\varepsilon_0})(\frac{q}{r})$

Where r is the radius of circular loop

Potential at a distance x from the centre of the loop is given by

$=(\frac{1}{4\pi\varepsilon_0})(\frac{q}{\sqrt{r^2+x^2}})$

Now Let takes A and B are respective centres of the two circular loop

Potential at A=Potential at A due to loop I + Potential due to Loop II

$V_A=(\frac{1}{4\pi\varepsilon_0})(\frac{q}{r}) + (\frac{1}{4\pi\varepsilon_0})(\frac{q}{\sqrt{r^2+x^2}})$

Now here x=.12m r=.09 and $q=10^{-6}$ C

Substituting these values

$V_A=2.4 \times 10^5$ V

Potential at B=Potential at B due to loop I + Potential due to Loop II

$V_B=(\frac{1}{4\pi\varepsilon_0})(\frac{q}{r}) + (\frac{1}{4\pi\varepsilon_0})(\frac{q}{\sqrt{r^2+x^2}})$

Now here x=.12m r=.05 and $q=10^{-6}$ C

So $V_B=1.7 \times 10^5$ V

So $V_A- V_B=7 \times 10^4$ V

**Question 5**
There are three charges on a straight line One Positive Charge q, Two Negative Charge -Q. Find the value of q/Q so that the entire system is in equilibrium? Will this equilibrium be stable?

Solution
For the charge q to be in equilibrium, the charges -Q should be at equal distance from it in opposite direction. For equilibrium of Charge Q, the sum of forces acting on it should also be zero. Let assume r be the distance between the charges

So

$\frac {Q^2}{4r^2} - \frac {Qq}{r^2}=0$

Hence q=Q/4

It does not depend on the distance r.

The equilibrium position is not stable. Since when charge -Q is shifted along left by a distance x, the force of attraction from positive charge

$F_1=\frac{Q^2}{4(a+x)^2}$

The force of repulsion from negative charge

$F_2=\frac{Q^2}{(2a+x)^2}$

It is clear that $F_2 > F_1$

So the charge Q will move still farther from the position of equilibrium

Similarly, if the charge -Q is move toward right the force of attraction will be more then Force of repulsion and it will move toward the centre.

Now if the charge q is moved right or left, the force of attraction on one side will more than the force of attraction of other side, so it will return to equilibrium position.

**Question 6**
A thread of length L placed along the x-axis with one end at the origin is electrified uniformly along its length with a net charge Q.Find the potential and electric field strength at a point P which is at a distance r from the origin along the axis of rod beyond the rod.

Solution
Consider a small elemental length dx having charge dq

Then $dq=(\frac {Q}{L})dx$.

Now potential due to this small charge at point P

$dV=K(\frac{Qdx}{Lx})$

Where $K=\frac{1}{4\pi\varepsilon_0}$

Potential at P due to total charge Q

$ V =\int_{r-l}^{r}\frac{KQdx}{Lx} = (\frac{KQ}{L})ln (\frac{r}{r-L})$

Electric Field is defined as

$E=-\frac{dV}{dr}=\frac{d}{dr} (\frac{KQ}{L})ln (\frac{r}{r-L})$

$=\frac{KQ}{Lr(r-L)}$

**Question 7**
A point electric dipole having dipole moment

**p **is placed in an external uniform electric field such that direction of dipole moment and electric field coincides. Find the radius of the sphere which forms one of the equipotential surfaces enclosing the dipole.

Solution
In an electric field say E potential at any point near the dipole is

$V=V_d-V_E$

Where $V_d$ is the potential at point P due to the dipole and $V_E$ is the potential at point P due to the uniform electric field.

If **r** is the position vector of point P from the centre of the dipole then

$V_d=\frac{k\mathrm{\mathbf{p}.} \hat{\mathrm{\mathbf{r}}}}{r^2}=\frac{k\mathrm{\mathbf{p}.\mathbf{r}} }{r^3}$

Now potential at P due to uniform electric field is

$V=\mathrm{\mathbf{E}.\mathbf{r}}$

$=\frac{k\mathrm{\mathbf{p}.\mathbf{r}} }{r^3} - \mathrm{\mathbf{E}.\mathbf{r}}= \frac {kpr \cos \theta}{r^3}-E r cos \theta$

$\theta$ is the angle which dipole axis makes with the uniform electric field. Thus,

$V=\left(\frac{kp}{r^3}-E\right)r \cos{\theta}$

For equipotential surface quantity in the brackets must be equal to zero.

$\Rightarrow\left(\frac{kp}{r^3}-E\right)=0$

or,

$=\left(\frac{kp}{E}\right)^{1/3}$

where,$k=\frac{1}{4\pi\varepsilon_0}$

is a constant.

**Question 8**
Calculate the potential and field due to a dipole of dipole moment 3.5 x 10

^{-11}C/m at a distance .5m from it

- on its axis
- on its perpendicular bisector

Solution
Electric potential due to electric dipole is given by

$V=\frac{1}{4\pi\varepsilon_0}\frac{pcos{\theta}}{r^2}$

(1) when point lies on the axis of the dipole

$V=\frac{1}{4\pi\varepsilon_0}\frac{p{cos0}^0}{r^2}=\frac{1}{4\pi\varepsilon_0}\frac{p}{r^2}$

$\frac{1}{4\pi\varepsilon_0}=9\times10^9$

Now putting the values of p and r in above equation we find

V=1.26 Volts

On axis of dipole electric field is given by

$E=\frac{2p}{4\pi\varepsilon_0r^3}$

Substituting in the required given values we find

E=2.52 V/m

(2) For points lying on the perpendicular bisector of the dipole $\theta=90^0$ thus,

$V=\frac{1}{4\pi\varepsilon_0}\frac{pcos{9}0^0}{r^2}=0$

And

$E=\frac{2p}{4\pi\varepsilon_0r^3}$

Substituting in the required given values we find

E=2.52 V/m

**Question 9**
15 identical mercury drops are charged to a same potential of 5 Volts. Assuming the drops to be of spherical shape, find the potential of the large drop, made up of the combination of all the charged drops.

Solution
Let r be the radius of each small drop and q be the amount of charge on each one of them. Potential at the surface of each drop is

$V=\frac{q}{4\pi\varepsilon_0r}$ Volt --- (1)

There are total 15 drops. Volume of the large drop would be same as the volume of all small drops. If R is the radius of the larger drop then,

$\frac{4}{3}\pi R^3=15\times\frac{4}{3}\pi r^3$

$ R=\left(15\right)^{1/3}r$

Charge on the large drop is Q=15q therefore potential of large drop is

$V'=\frac{15q}{4\pi\varepsilon_0(15)^{1/3}r}$ Volt

$=\frac{15V}{(15)^{1/3}}$

Given that V=5 Volt substituting the value in above equation and calculating we find

V'=30.4Volts

**Question 10**
Two charges -q each are separated by distance 2d. A third charge + q is kept at mid point O. Find potential energy of + q as a function
of small distance x from O due to -q charges. Sketch P.E. v/s x and convince yourself that the charge at O is in an unstable equilibrium

Solution
Potential energy is given by

$U= \frac {1}{4 \pi \epsilon _0}[ \frac {-q^2}{d+x} + \frac {-q^2}{d-x}]$

or

$U= \frac {-q^2}{4 \pi \epsilon _0}\frac {2d}{d^2 -x^2}$

Sketching the graph between P.E. v/s x

$\frac {dU}{dx}= \frac {-2dq^2}{4 \pi \epsilon _0}\frac {2x}{(d^2 -x^2)^2}$

Now

$\frac {dU}{dx}=0$ at x=0

x = 0 is an equilibrium point.

$U_0= \frac {2q^2}{4 \pi \epsilon _0 d}$

Now doing double differentiation and checking the point of maxima and minima

$\frac {d^2U}{dx^2} = \frac {-2dq^2}{4 \pi \epsilon _0} [ \frac {2}{(d^2 -x^2)^2} - \frac {8x^2}{(d^2 -x^2)^3}]$

or

$=(\frac {-2dq^2}{4 \pi \epsilon _0} )(\frac {1}{(d^2 -x^2)^3}) [2 (d^2 -x^2)^2 -8x^2]$

At x=0, we have

$\frac {d^2U}{dx^2} = (\frac {-2dq^2}{4 \pi \epsilon _0} )(\frac {1}{d^6})(2d^2) $

The above quantity is less than zero, so this is the point of maxima and hence unstable equlibrium

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Class 12 Maths
Class 12 Physics