JEE Main 2023 — Electrostatics Question with Solution

Given two charges each of magnitude $$0.01\text{C}$$ and separated by a distance of $$0.4\text{mm}=0.4\times 10^{-3}\text{m}$$, we can calculate the dipole moment $$\vec{p}$$:

$$\vec{p}=q\cdot \vec{d}=(0.01\text{C})(0.4\times 10^{-3}\text{m})=4\times 10^{-6}\text{Cm}$$

The dipole is placed in a uniform electric field $$\vec{E}$$ of magnitude $$10\text{dyne/C}$$, which is equivalent to $$10\times 10^{-5}\text{N/C}$$.

The angle between the dipole moment and the electric field is $$30^{\circ }$$.

The torque acting on the dipole can be calculated using the formula:

$$\tau =pE\sin \theta$$

Substituting the given values:

$$\tau =(4\times 10^{-6}\text{Cm})(10\times 10^{-5}\text{N/C})\sin 30^{\circ }$$

$$\tau =(4\times 10^{-6}\text{Cm})(10^{-4}\text{N/C})(\frac{1}{2})$$

$$\tau =2\times 10^{-10}\text{Nm}$$

The magnitude of the torque acting on the dipole is $$2.0\times 10^{-10}\text{Nm}$$.