Electrostatics

JEE Main level, single correct, medium

235 of 311

Question 235 of 311

A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an electric field E⃗1=E0x^\vec{E}_1=E_0\hat{x}. If another electric field E⃗2=2E0(y^+z^)\vec{E}_2=2E_0(\hat{y}+\hat{z}) is introduced to the system, what will be the percentage change in the frequency of the oscillation (approximate)?

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Answer: (A) 73%73\%

The frequency of oscillation of a rigid dipole in an electric field is given by f=12πpEIf = \dfrac{1}{2\pi} \sqrt{\dfrac{pE}{I}}, where pp is the dipole moment, EE is the magnitude of the electric field, and II is the moment of inertia.

This implies f∝Ef \propto \sqrt{E}.

The initial electric field is E⃗1=E0x^\vec{E}_1 = E_0 \hat{x}, so its magnitude is E1=E0E_1 = E_0.

The initial frequency is f1∝E0f_1 \propto \sqrt{E_0}.

When the second electric field E⃗2=2E0(y^+z^)\vec{E}_2 = 2E_0(\hat{y} + \hat{z}) is introduced, the net electric field becomes E⃗net=E⃗1+E⃗2=E0x^+2E0y^+2E0z^\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 Enet=E02+(2E0)2+(2E0)2=E02+4E02+4E02=3E0E_{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 f2∝3E0f_2 \propto \sqrt{3E_0}.

The percentage change in the frequency is given by f2−f1f1×100%\dfrac{f_2 - f_1}{f_1} \times 100\%.

Substituting the values, we get (3E0E0−1)×100%=(3−1)×100%\left( \dfrac{\sqrt{3E_0}}{\sqrt{E_0}} - 1 \right) \times 100\% = (\sqrt{3} - 1) \times 100\%.

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

The approximate percentage change is 73%73\%.

Answer: 73%73\%

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