Question

For two points $A$ and $B$ in an electric field, if $\Delta W$ is work done in taking a test charge $q^{\prime }$ 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 $AB$ from $A$ to $B$ and $ds$ is differential displacement along path $AB$.
Then which of the following options is correct?

• A. $V_B-V_A=\underset{A}{\overset{B}{\int }}E\cdot ds$
• B. $\Delta W=-q^{\prime }\underset{A}{\overset{B}{\int }}E\cdot ds$
• C. $\Delta V=\frac{\Delta U}{q^{\prime }}$
• D. $\Delta V=\frac{\Delta W}{q^{\prime }}=\frac{\Delta U}{q^{\prime }}$

Answer 1

Answer: (C)

For small displacement $ds$, work done by the electric field $(\Delta W{)}_{\text{electric force }}=F_e\cdot dr$
The change in potential $\Delta V$ is given
$\Delta V=\frac{-[\Delta W{]}_{\text{electric force }}}{q^{\prime }}$
$=\frac{\Delta U}{q^{\prime }}$