An electric dipole consists of two particles, | vedclass

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(A) The dipole moment is $p = q 	imes d = (1 	imes 10^{-6} \ C) 	imes (1 \ m) = 10^{-6} \ Cm$. The moment of inertia $I$ of the system about the ce

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$2.5$

Vedclass · Answer

(A) The dipole moment is $p = q 	imes d = (1 	imes 10^{-6} \ C) 	imes (1 \ m) = 10^{-6} \ Cm$. The moment of inertia $I$ of the system about the center is $I = m(d/2)^2 + m(d/2)^2 = m(d^2/2) = 1 	imes (1^2/2) = 0.5 \ kg \ m^2$. For a small angular displacement $	heta$, the restoring torque is $	au = -pE \sin 	heta \approx -pE 	heta$. Since $	au = I \alpha$, we have $I \alpha = -pE 	heta$, which gives $\alpha = -(pE/I) 	heta$. This represents simple harmonic motion with angular frequency $\omega = \sqrt{pE/I}$. Substituting the values: $\omega = \sqrt{(10^{-6} 	imes 2 	imes 10^4) / 0.5} = \sqrt{2 	imes 10^{-2} / 0.5} = \sqrt{0.04} = 0.2 \ rad/s$. The time taken to return to the mean position from the extreme position is $t = T/4 = (2 \pi / \omega) / 4 = \pi / (2 \omega)$. $t = \pi / (2 	imes 0.2) = \pi / 0.4 = 2.5 \pi \ s$.

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