Two charged particles each of mass 9.8 text{ | vedclass

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(A) The arrangement of the electric dipole in the electric field will undergo Simple Harmonic Motion $(SHM)$ when the rod is set free. The time period

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$5$

Vedclass · Answer

(A) The arrangement of the electric dipole in the electric field will undergo Simple Harmonic Motion $(SHM)$ when the rod is set free. The time period $T$ of this $SHM$ is given by:$T = 2\pi \sqrt{\frac{I}{pE}}$where $I$ is the moment of inertia of the dipole,$p$ is the electric dipole moment,and $E$ is the electric field.The moment of inertia $I$ about the center of the rod is $I = m(l/2)^2 + m(l/2)^2 = 2m(l/2)^2 = \frac{ml^2}{2}$.Given: $m = 9.8 	imes 10^{-3} 	ext{ kg}$,$l = 0.5 	ext{ m}$,$q = 20 	imes 10^{-6} 	ext{ C}$,$E = 12.1 	ext{ N/C}$.$I = \frac{9.8 	imes 10^{-3} 	imes (0.5)^2}{2} = 4.9 	imes 10^{-3} 	imes 0.25 = 1.225 	imes 10^{-3} 	ext{ kg m}^2$.$p = q 	imes l = 20 	imes 10^{-6} 	imes 0.5 = 10^{-5} 	ext{ C m}$.$T = 2\pi \sqrt{\frac{1.225 	imes 10^{-3}}{10^{-5} 	imes 12.1}} = 2\pi \sqrt{\frac{1.225 	imes 100}{12.1}} = 2\pi \sqrt{\frac{122.5}{12.1}} \approx 2\pi \sqrt{10.12} \approx 2\pi 	imes 3.18 \approx 20 	ext{ s}$.The time taken by the rod to move from the initial small angle to the parallel position (equilibrium position) is $T/4$.Required time $= \frac{20}{4} = 5 	ext{ s}$.

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