A particle of mass 100, g and charge 2, mu C | vedclass

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(A) The principle of conservation of energy states that the loss in electrostatic potential energy is equal to the gain in kinetic energy of the parti

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(A) The principle of conservation of energy states that the loss in electrostatic potential energy is equal to the gain in kinetic energy of the particle.Initial potential energy $U_1 = \frac{kQq}{r_1}$ and final potential energy $U_2 = \frac{kQq}{r_2}$.Change in potential energy $\Delta U = U_1 - U_2 = kQq \left( \frac{1}{r_1} - \frac{1}{r_2} ight)$.Given: $m = 0.1\, kg$,$q = 2 	imes 10^{-6}\, C$,$Q = 5 	imes 10^{-6}\, C$,$r_1 = 0.5\, m$,$r_2 = 3\, m$,$k = 9 	imes 10^9\, N\cdot m^2/C^2$.$\Delta U = (9 	imes 10^9) 	imes (5 	imes 10^{-6}) 	imes (2 	imes 10^{-6}) 	imes \left( \frac{1}{0.5} - \frac{1}{3} ight) = 0.09 	imes (2 - 0.333) = 0.09 	imes 1.666 = 0.15\, J$.Since $\Delta U = \Delta K = \frac{1}{2}mv^2$,we have $0.15 = \frac{1}{2} 	imes 0.1 	imes v^2$.$v^2 = \frac{0.15 	imes 2}{0.1} = 3$.$v = \sqrt{3} \approx 1.73\, m/s$.

$A$ particle of mass $100\, g$ and charge $2\, \mu C$ is released from a distance of $50\, cm$ from a fixed charge of $5\, \mu C$. Find the speed of the particle when its distance from the fixed charge becomes $3\, m$. Neglect any other force. (Result in $m/s$) (in $.73$)

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