Question

Two concentric conducting spheres of radii $R$ and $2R$ are carrying charges $Q$ and $2Q$ respectively. If the charge on inner sphere is doubled, the potential difference between the two spheres will :

• A. become two times
• B. become four times
• C. be halved
• D. remain same

Answer 1

The correct option is A become two times

Potential difference between the two spheres $\Delta V=V_A-V_B$

Initially :

$V_A=\frac{kQ}{R}-\frac{k\left(2Q\right)}{2R}=0$

where $k=\frac{1}{4\pi {ϵ}_o}$

$V_B=\frac{kQ}{2R}-\frac{k\left(2Q\right)}{2R}=\frac{-kQ}{2R}$

$⟹$ $\Delta V_i=\frac{kQ}{2R}$

Now the charge on the inner sphere is doubled.

$V_A=\frac{k\left(2Q\right)}{R}-\frac{k\left(2Q\right)}{2R}=\frac{kQ}{R}$

$V_B=\frac{k\left(2Q\right)}{2R}-\frac{k\left(2Q\right)}{2R}=0$

$\therefore$ $\Delta V_f=\frac{KQ}{R}-0$

$⟹$ $\Delta V_f=\frac{kQ}{R}$

We get $\Delta V_f=2\Delta V_i$

Hence potential difference between the two spheres will become two times.