Question

Figure shows three spherical and equipotential surfaces A, B and C round a point charge q. The potential difference $V_A - V_B = V_B - V_C. \, If \, t_1 \, and \, t_2$ be the distances between them, then

• A. $ t_1 = t_1$
• B. $ t_1 > t_2$
• C. $ t_1 < t_2$
• D. $ t_1 \le t_2$

Answer 1

Answer: (C)

PotentiaI difference between two equipotential surfaces A and B.
$ V_A - V_B = kq \bigg( \frac{ 1}{ r_A } - \frac{ 1}{ r_B} \bigg) = kq \bigg( \frac{ r_B - r_A }{ r_A \, r_B} \bigg) = \frac{ kq t_1 }{ r_A \, r_B}$
or $ t_1 = \frac{ ( V_A - V_B) \, r_A \, r_B}{ kq}$
or $ t_1 \propto r_A \, r_B$
Simiiarly, $ t_1 \propto r_B \, r_C$
Since, $ r_A < r_B < r_C, \, therefore \, r_A \, r_B < r_B \, r_C$
$\therefore t_1 < t_2 $