Question

A conducting sphere A having a radius a is charged to a potential $V_1$. Now it is surrounded by another conducting spherical shell of radius b. The potential $V_2$ acquired by the sphere A after it is connected to shell B by a thin conducting wire is

• A. $V_1\frac{a}{b}$
• B. $V_1\frac{b}{a}$
• C. $\frac{V_1(a+b)}{b}$
• D. $\frac{V_{1}a}{a+b}$

Answer 1

The correct option is D

$\frac{V_{1}a}{a+b}$

Let the initial charge on A is $q$.

$V_1=\frac{kq}{a}$ .......(1)

Now, when wire is connected between A and B, charges will flow from A to B till they attain an equal potential.

Let the new charge on A be $q^{\prime }$.

Hence the charge on B is $q-q^{\prime }$.

So, $V_2=\frac{k{q}^{\prime }}{a}=\frac{k(q-q^{\prime })}{b}$........(2)

Solving (1) and (2), we get :

$V_2=\frac{V_{1}a}{a+b}$