Question

In moving from $A$ to $B$ along an electric field line, the work done by the electric field on an electron is $6.4\times {10}^{-19}J$. If ${\varphi }_1$ and ${\varphi }_2$ are equipotential surfaces, then the potential difference $V_c-V_A$ is :

• A. $-4V$
• B. $4V$
• C. $0$
• D. $6.4V$

Answer 1

Answer: (B)

$4V$

Work done will be equal to change in potential energy
$W_{eL}=q(V_i-V_f)$
or $6.4\times {10}^{-19}=-1.6\times {10}^{-19}(V_A-V_B)$
or $V_A-V_B=-4V$
or $V_A-V_C=-4V(\because V_B=V_C)$
or $V_C-V_A=4V$