Question

For two points $A$ and $B$ in an electric field, if $\Delta W$ is work done in taking a test charge $q'$ from $A$ to $B, \Delta U=V_{B}-V_{A}$ is change in potential energy of a test charge when it is moved from $A$ to $B$.

$\Delta V=V_{B}-V_{A}$ is change in potential due to field if we move along path $A B$ from $A$ to $B$ and $d s$ is differential displacement along path $A B$.
Then which of the following options is correct?

• A. $V_{B}-V_{A}=\int\limits_{A}^{B} E \cdot d s$
• B. $\Delta W=-q'\int\limits_{A}^{B} E \cdot d s$
• C. $\Delta V=\frac{\Delta U}{q'}$
• D. $\Delta V=\frac{\Delta W}{q'}=\frac{\Delta U}{q'}$

Answer 1

Answer: (C)

For small displacement $d s$, work done by the electric field $(\Delta W)_{\text {electric force }}=F_{e} \cdot d r$
The change in potential $\Delta V$ is given
$\Delta V=\frac{-[\Delta W]_{\text {electric force }}}{q'}$
$=\frac{\Delta U}{q'}$