Two conducting concentric , hollow sphere A and B have radii a and b respectively with A inside B . Their common potentials is V . A is now given some charge such that its potential becomes zero . The potential of B will now be

Question

$(a)\;0\qquad(b)\;V (1-\large\frac{a}{b})\qquad(c)\;\large\frac{V_{a}}{b}\qquad(d)\;\large\frac{V_{b}}{a}$

Answer 1

Answer : (b) $\;V (1-\large\frac{a}{b})$

Explanation :

Let the charge on sphere A is $\;q_{A}\;$ charge on sphere B is $\;q_{B}\;$ initially.

Then , $\;V_{A}=\large\frac{k\;q_{A}}{a}+\large\frac{k\;q_{B}}{b}$

and $\;V_{B}=\large\frac{k\;q_{A}}{b}+\large\frac{k\;q_{B}}{b}$

But $\;V_{A}=V_{B}=V \quad\;\large\frac{k\;q_{A}}{a}+\large\frac{k\;q_{B}}{b}=\large\frac{k\;q_{A}}{b}+\large\frac{k\;q_{B}}{b}$

$q_{A}=0$

Now let the charge on A becomes q such that ite potential becomes zero

Then $\;V_{A}=\large\frac{k\;q}{a}+\large\frac{k\;q_{B}}{b}=0$

$q=-\large\frac{q_{B a}}{b}$

$V_{B}=\large\frac{k\;q}{b}+\large\frac{k\;q_{B}}{b}=\large\frac{k\;q_{B}}{b}-\large\frac{k\;q_{B}\;a}{b^2}$

$V_{B}=\large\frac{k\;q_{B}}{b}\;(1-\large\frac{a}{b})$

By $\;\large\frac{k\;q_{B}}{b}=V$

Therfore $\;V_{B}=V\;(1-\large\frac{a}{b})\;.$