Question

Two concentric shells have radii R and 2R, charges $q_A$ and $q_B$ and potentials 2V and (3/2)V respectively. Now shell B is earthed and let charges on them become $q_A^{\prime }$ and $q_B^{\prime }$. Then:

• A. $q_A/q_B=1/2$
• B. Potential of A after earthing becomes (3/2)V
• C. $|q_A^{\prime }|/|q_B^{\prime }|=1$
• D. Potential difference between A and B after earthing becomes V/2

Answer 1

Answer: (A), (C), (D)

Potential of A:

$2V=\frac{q_A}{R}+\frac{q_B}{2R}$

Potential of B:

$\frac{3V}{2}=\frac{q_A}{2R}+\frac{q_B}{2R}$

Solving both,

$q_A=VR$

$q_B=2VR$

$\frac{q_A}{q_B}=\frac{1}{2}$

When shell B is earthed, $V_B^{\prime }=0$

$\Rightarrow \frac{q_A^{\prime }}{2R}+\frac{q_B^{\prime }}{2R}=0$

$\Rightarrow q_B^{\prime }=-q_A^{\prime }$

$\frac{|q_A^{\prime }|}{|q_B^{\prime }|}=1$

$V_A^{\prime }=\frac{q_A^{\prime }}{R}+\frac{q_B^{\prime }}{2R}=\frac{q_A^{\prime }}{2R}$

Potential difference between A and B: $V_A^{\prime }-V_B^{\prime }=V_A-V_B=2V-\frac{3}{2}V=\frac{1}{2}V$

$\Rightarrow V_A^{\prime }=\frac{1}{2}V$ since $V_B^{\prime }=0$