Question

Two small identical metal balls $A$ and $B$ of 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 charges on A and B are (given $r<<a_0)$

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

Answer 1

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

Given that $r<<a_0$. Thus, the charge distribution on the sphere doesn't change due to induction.

The potential due to a spherical shell of charge is given by $V=\frac{Q}{4\pi {ϵ}_{0}r}$ for $r\ge R$

Let $q_1$ and $q_2$ be the charges on each of the spheres.

Then, potential on the surface of first sphere is $V_1=\frac{q_1}{4\pi {ϵ}_{0}r}+\frac{q_2}{4\pi {ϵ}_0{a}_0}$

Similarly, the potential on second sphere is $V_2=\frac{q_1}{4\pi {ϵ}_0{a}_0}+\frac{q_2}{4\pi {ϵ}_{0}r}$

The two equations can be written as

$a_0{q}_1+r{q}_2=4\pi {ϵ}_0{a}_{0}r{V}_1-----\left(1\right)r{q}_1+a_0{q}_2=4\pi {ϵ}_0{a}_{0}r{V}_2-----\left(2\right)$

Now $r\times \left(1\right)-a_0\times \left(2\right)$ give us

$\Rightarrow (r^2-a_0^2)q_2=4\pi {ϵ}_0{a}_{0}r(r{V}_1-a_0{V}_2)\Rightarrow q_2=4\pi {ϵ}_0\frac{a_{0}r(-V_{1}r+V_2{a}_0)}{a_0^2-r^2}$

Similarly, $a_0\times \left(1\right)-r\times \left(2\right)$ give us

$\Rightarrow (a_0^2-r^2)q_1=4\pi {ϵ}_0{a}_{0}r(a_0{V}_1-r{V}_2)\Rightarrow q_1=4\pi {ϵ}_0\frac{a_{0}r(V_1{a}_0-V_{2}r)}{a_0^2-r^2}$