A particle of mass 1 text{ g} and charge 1 mu | vedclass

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(D) The work done by the electrostatic force on the particle is equal to the change in its kinetic energy.Work done $W = \int_{r_1}^{r_2} \frac{k q_1

Vedclass · Accepted Answer

$180$

Vedclass · Answer

(D) The work done by the electrostatic force on the particle is equal to the change in its kinetic energy.Work done $W = \int_{r_1}^{r_2} \frac{k q_1 q_2}{r^2} dr = k q_1 q_2 \left[ -\frac{1}{r} ight]_{1}^{10} = k q_1 q_2 \left( 1 - \frac{1}{10} ight) = k q_1 q_2 \left( \frac{9}{10} ight)$.Given $k = 9 	imes 10^9 	ext{ N m}^2 	ext{ C}^{-2}$, $q_1 = 1 	imes 10^{-6} 	ext{ C}$, $q_2 = 2 	imes 10^{-3} 	ext{ C}$, $m = 1 	imes 10^{-3} 	ext{ kg}$.$W = (9 	imes 10^9) 	imes (1 	imes 10^{-6}) 	imes (2 	imes 10^{-3}) 	imes 0.9 = 18 	imes 0.9 = 16.2 	ext{ J}$.Using the work-energy theorem: $W = \frac{1}{2} m v^2$.$16.2 = \frac{1}{2} 	imes (1 	imes 10^{-3}) 	imes v^2$.$v^2 = \frac{16.2 	imes 2}{10^{-3}} = 32.4 	imes 10^3 = 32400$.$v = \sqrt{32400} = 180 	ext{ m s}^{-1}$.

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