Physics - Electrostatics Question with Solution | TestHub

The electric potential due to a dipole $\vec{p}$ at a position $\vec{r}$ is $V=\frac{k\vec{p}\cdot \vec{r}}{r^3}$. For point A $(a,0)$, ${\vec{r}}_A=a\hat{i}$. $V_A=\frac{k(p\hat{i})\cdot (a\hat{i})}{a^3}=\frac{kpa}{a^3}=\frac{kp}{a^2}$. For point B $(0,a)$, ${\vec{r}}_B=a\hat{j}$. $V_B=\frac{k(p\hat{i})\cdot (a\hat{j})}{a^3}=0$. Work done by electric field $W_E=q(V_A-V_B)=q(\frac{kp}{a^2}-0)=\frac{kpq}{a^2}$. So, option A is correct. Change in potential energy $\Delta U=U_B-U_A=q(V_B-V_A)=q(0-\frac{kp}{a^2})=-\frac{kpq}{a^2}$. This means the potential energy decreases by $\frac{kpq}{a^2}$. So, option B is correct. By work-energy theorem, $K_B-K_A=W_E$. If released from rest, $K_A=0$. So $K_B=W_E=\frac{kpq}{a^2}$. So, option C is correct.

The electric field at an equatorial point $(0,a)$ for a dipole along the x-axis is ${\vec{E}}_B=-\frac{k\vec{p}}{a^3}=-\frac{kp\hat{i}}{a^3}$, which is not zero. So, option D is incorrect.