Question

two concentric conducting spheres of radii R and 2r are carrying charges q and minus 2 respectively if the charge on inner sphere is doubled the potential difference between the two spheres will be​

Answer 1

The potential difference between the two spheres will be  $\bold{V_2=2V_1}$

Given:

Charge = q1 = q C

Charge = q2 = -2q C

To find:

Potential difference between the two spheres

Formula:

Potential difference of a sphere is given by the formula:

$\bold{V= \frac{kq}{r}}$

Where,

q = Charge

r = Radius

Solution:

Potential difference between sphere A and sphere B:

$\bold{V_1=(\frac{kq}{R}-\frac{k2q}{2r})-(\frac{kq}{2r}-\frac{k2q}{2r})}$

$\bold{V_1=\frac{kq}{R}-\frac{k2q}{2r}+\frac{kq}{2r}}$

$\bold{\therefore V_1=\frac{kq}{R}-\frac{kq}{2r}}$

After being doubled, potential difference between sphere A and sphere B:

$\bold{V_2=(\frac{k2q}{R}-\frac{k2q}{2r})-(\frac{k2q}{2r}-\frac{k2q}{2r})}$

$\bold{V_2=\frac{k2q}{R}-\frac{k2q}{2r}}$

$\bold{V_2=2(\frac{kq}{R}-\frac{kq}{2r})}$

$\bold{\therefore V_2=2V_1}$