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

The frequency of oscillation of a rigid dipole in an electric field is given by \(f = \dfrac{1}{2\pi} \sqrt{\dfrac{pE}{I}}\), 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 \(\dfrac{f_2 - f_1}{f_1} \times 100\%\).

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

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

The approximate percentage change is \(73\%\).

Answer: \(73\%\)