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(B) The moment of inertia $I$ of the dipole about its center is given by:$I = m\left(\frac{d}{2}\right)^2 + m\left(\frac{d}{2}\right)^2 = 2 \cdot m \f

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$\sqrt{\frac{2qE}{md}}$

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(B) The moment of inertia $I$ of the dipole about its center is given by:$I = m\left(\frac{d}{2}ight)^2 + m\left(\frac{d}{2}ight)^2 = 2 \cdot m \frac{d^2}{4} = \frac{md^2}{2}$The torque $	au$ acting on the dipole in a uniform electric field $E$ is given by:$	au = pE \sin 	heta = (qd) E \sin 	heta$Using the rotational analog of Newton's second law,$	au = I \alpha$,where $\alpha$ is the angular acceleration:$(qEd) \sin 	heta = \left(\frac{md^2}{2}ight) \alpha$For small angles $	heta$,$\sin 	heta \approx 	heta$. Therefore:$(qEd) 	heta = \left(\frac{md^2}{2}ight) \alpha$$\alpha = \left(\frac{2qE}{md}ight) 	heta$Comparing this with the equation for simple harmonic motion $\alpha = \omega^2 	heta$,we get:$\omega^2 = \frac{2qE}{md}$$\omega = \sqrt{\frac{2qE}{md}}$

An electric dipole is formed by two equal and opposite charges $q$ with separation $d$. The charges have the same mass $m$. It is kept in a uniform electric field $E$. If it is slightly rotated from its equilibrium orientation,then its angular frequency $\omega$ is

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