JEE Main 2026 — Physics Electromagnetism

The frequency of oscillation of a rigid dipole in an electric field is given by $f = \dfrac{1}{2\pi} \sqrt{\dfrac{pE}{I}}$f=2π1​IpE​​, where $p$p is the dipole moment, $E$E is the magnitude of the electric field, and $I$I is the moment of inertia.

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

The initial electric field is $\vec{E}_1 = E_0 \hat{x}$E1​=E0​x^, so its magnitude is $E_1 = E_0$E1​=E0​.

The initial frequency is $f_1 \propto \sqrt{E_0}$f1​∝E0​​.

When the second electric field $\vec{E}_2 = 2E_0(\hat{y} + \hat{z})$E2​=2E0​(y^​+z^) is introduced, the net electric field becomes $\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 = E_0 \hat{x} + 2E_0 \hat{y} + 2E_0 \hat{z}$Enet​=E1​+E2​=E0​x^+2E0​y^​+2E0​z^.

The magnitude of the net electric field is $E_{net} = \sqrt{E_0^2 + (2E_0)^2 + (2E_0)^2} = \sqrt{E_0^2 + 4E_0^2 + 4E_0^2} = 3E_0$Enet​=E02​+(2E0​)2+(2E0​)2​=E02​+4E02​+4E02​​=3E0​.

The new frequency is $f_2 \propto \sqrt{3E_0}$f2​∝3E0​​.

The percentage change in the frequency is given by $\dfrac{f_2 - f_1}{f_1} \times 100\%$f1​f2​−f1​​×100%.

Substituting the values, we get $\left( \dfrac{\sqrt{3E_0}}{\sqrt{E_0}} - 1 \right) \times 100\% = (\sqrt{3} - 1) \times 100\%$(E0​​3E0​​​−1)×100%=(3​−1)×100%.

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

The approximate percentage change is $73\%$73%.

Answer: $73\%$73%