Q. In moving from $A$ to $B$ along an electric field line, the electric field does $6.4 \times 10^{- 19}J$ of work on an electron. If $\phi_{1},\phi_{2}$ are equipotential surfaces, then the potential difference $\left( V_{C} - V_{A} \right)$ is
A) $- 4\ V$
B) $4\ V$
C) Zero
D) $64\ V$
Explanation
(b) Work done by the field $W = q( - dV) = - e(V_{A} - V_{B})$
$= e\left( V_{B} - V_{A} \right) = e(V_{C} - V_{A})$ $\lbrack\because V_{B} = V_{C}\rbrack$
$\Rightarrow \left( V_{C} - V_{A} \right) = \frac{W}{e} = \frac{6.4 \times 10^{- 19}}{1.6 \times 10^{- 19}} = 4V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.