Question

Two small identical metal balls A and B radius r are placed apart The distance between centre of balls is $a_0$ The net potential of ball A is $V_1$ and that of B is $V_2$ Let $q_1$ and $q_2$ are the charges on balls A and B respectively Then the charge on A and B are ($(givenr<<a_0)$)

• A. $q_1=4\pi {\epsilon }_0\frac{a_{0}r(V_1{a}_0-V_{2}r)}{a_0^2-r^2}$
• B. $q_1=4\pi {\epsilon }_0\frac{a_{0}r(V_1{a}_0+V_{2}r)}{a_0^2+r^2}$
• C. $q_2=4\pi {\epsilon }_0\frac{a_{0}r(V_2{a}_0+V_{1}r)}{a_0^2+r^2}$
• D. $q_2=4\pi {\epsilon }_0\frac{a_{0}r(V_2{a}_0-V_{1}r)}{a_0^2-r^2}$

Answer 1

Answer: (A), (D)

The correct options are
A $q_1=4\pi {\epsilon }_0\frac{a_{0}r(V_1{a}_0-V_{2}r)}{a_0^2-r^2}$
D $q_2=4\pi {\epsilon }_0\frac{a_{0}r(V_2{a}_0-V_{1}r)}{a_0^2-r^2}$

The potential on Ball A is:

$V_1=\frac{k{q}_1}{r}+\frac{k{q}_2}{a_0}$ ...(i)

i.e, potential is sum of potential on the ball A due to charge $q_1$ and also due to charge $q_2$ on ball B at a distance $a_0$ apart.

Similarly, the potential on ball B is,

$V_2=\frac{k{q}_2}{r}+\frac{k{q}_1}{a_0}$ ...(ii)

where $k=\frac{1}{4\pi {\epsilon }_0}$

From (i),

$q_1=\frac{r}{k}(V_1-\frac{k{q}_2}{a_0})$ ...(iii)

Substituting in (ii), we get:

$V_2=\frac{k{q}_2}{r}+\frac{k}{a_0}\frac{r}{k}[V_1-\frac{k{q}_2}{a_0}]$

On solving for $q_2$, we get:

$q_2=\frac{1}{k}{a}_{0}r\frac{(V_2{a}_0-V_{1}r)}{a_0^2-r^2}$

$q_2=\frac{4\pi {\epsilon }_0{a}_{0}r}{(a_0^2-r^2)}[V_2{a}_0-V_{1}r]$

Substituting in (iii) we get,

$q_1=\frac{r}{k}[V_1-\frac{k}{a_0}\left(\frac{1}{k}\frac{a_{0}r(V_2{a}_0-V_{1}r)}{a_0^2-r^2}\right)]$

$=\frac{r}{k}{V}_1-\frac{r^2}{k}\left(\frac{V_2{a}_0-V_{1}r}{a_0^2-r^2}\right)$

$=\frac{V_1}{k}[r+\frac{r^3}{a_0^2-r^2}]+\frac{V_2{r}^2}{k}[r+\frac{a_0}{a_0^2-r^2}]$

On solving, we get

$q_1=4\pi {\epsilon }_0\frac{a_{0}r(V_1{a}_0-V_{2}r)}{a_0^2-r^2}$