Physics - W.E.P. Question with Solution | TestHub

The electric potential at a distance $r$ from a point charge $Q$ is given by $V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$.

From the given figure, both points $A$ and $B$ are at the same distance $r$ from the charge $+Q$.

Therefore, the electric potential at point $A$ is $V_A=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$ and the electric potential at point $B$ is $V_B=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$.

Since $V_A=V_B$, the potential difference between points $A$ and $B$ is $\Delta V=V_B-V_A=0$.

The work done in moving a charge $q$ from $A$ to $B$ is $W=q(V_B-V_A)=q(0)=0$.

Alternatively, the path $AB$ is a circular arc centered at $+Q$, which means the electric force on charge $q$ is always radial and perpendicular to the tangential displacement. Since the angle between force and displacement is $90^{\circ }$ at every point, the work done is zero.

Answer: zero