Q.
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 distance between them. Then
A) $t_{1} = t_{2}$
B) $t_{1} > t_{2}$
C) $t_{1} < t_{2}$
D) $t_{1} \leq t_{2}$
Explanation
(c) Potential difference between two equipotential surfaces A and B.
$V_{A} - V_{B} = kq\ \left( \frac{1}{r_{A}} - \frac{1}{r_{B}} \right)$
$= kq\left( \frac{r_{B} - r_{A}}{r_{A}r_{B}} \right)$
$= \frac{kqt_{1}}{r_{A}r_{B}}$
Or
$t_{1} = \frac{\left( V_{A} - V_{B} \right)r_{A}r_{B}}{kq}$
Or $t_{1} \propto r_{A}r_{B}$
Similarly, $t_{2} \propto r_{B}r_{C}$
Since,
$r_{A} < r_{B} < r_{C},$ therefore $r_{A}r_{B} < r_{B}r_{C}\ $
$\therefore\ \ t_{1} < t_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.