JEE Main 2026 — Electrostatics Question with Solution

The frequency of oscillation of a rigid dipole in an electric field is given by $f=\frac{1}{2\pi }\sqrt{\frac{pE}{I}}$, where $p$ is the dipole moment, $E$ is the magnitude of the electric field, and $I$ is the moment of inertia.

This implies $f\propto \sqrt{E}$.

The initial electric field is ${\vec{E}}_1=E_0\hat{x}$, so its magnitude is $E_1=E_0$.

The initial frequency is $f_1\propto \sqrt{E_0}$.

When the second electric field ${\vec{E}}_2=2E_0(\hat{y}+\hat{z})$ is introduced, the net electric field becomes ${\vec{E}}_{net}={\vec{E}}_1+{\vec{E}}_2=E_0\hat{x}+2E_0\hat{y}+2E_0\hat{z}$.

The magnitude of the net electric field is $E_{net}=\sqrt{E_0^2+(2E_0{)}^2+(2E_0{)}^2}=\sqrt{E_0^2+4E_0^2+4E_0^2}=3E_0$.

The new frequency is $f_2\propto \sqrt{3E_0}$.

The percentage change in the frequency is given by $\frac{f_2-f_1}{f_1}\times 100\%$.

Substituting the values, we get $\left(\frac{\sqrt{3E_0}}{\sqrt{E_0}}-1\right)\times 100\%=(\sqrt{3}-1)\times 100\%$.

Using $\sqrt{3}\approx 1.732$, the percentage change is $(1.732-1)\times 100\%=73.2\%$.

The approximate percentage change is $73\%$.

Answer: $73\%$