A particle of mass 0.2 gm and charge 1 mu C | vedclass

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(B) Given: Mass $m = 0.2 \ gm = 0.2 	imes 10^{-3} \ kg$,Charge $q_1 = 1 \ \mu C = 10^{-6} \ C$,Charge $q_2 = 1 \ \mu C = 10^{-6} \ C$.Initial distanc

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$9$

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(B) Given: Mass $m = 0.2 \ gm = 0.2 	imes 10^{-3} \ kg$,Charge $q_1 = 1 \ \mu C = 10^{-6} \ C$,Charge $q_2 = 1 \ \mu C = 10^{-6} \ C$.Initial distance $r_1 = 1 \ m$,Final distance $r_2 = 10 \ m$.By the principle of conservation of energy,the loss in electrostatic potential energy is equal to the gain in kinetic energy:$\Delta K = -\Delta U$$\frac{1}{2} m v^2 = \frac{K q_1 q_2}{r_1} - \frac{K q_1 q_2}{r_2}$$\frac{1}{2} (0.2 	imes 10^{-3}) v^2 = (9 	imes 10^9) (10^{-6}) (10^{-6}) \left( \frac{1}{1} - \frac{1}{10} ight)$$0.1 	imes 10^{-3} v^2 = 9 	imes 10^{-3} 	imes (0.9)$$10^{-4} v^2 = 8.1 	imes 10^{-3}$$v^2 = 81$$v = 9 \ m/s$.

$A$ particle of mass $0.2 \ gm$ and charge $1 \ \mu C$ is held at a distance of $1 \ m$ from a fixed charge of $1 \ \mu C$. If the particle is released,it will be repelled. The speed of the particle when it is at a distance of $10 \ m$ from the fixed charge is $.... \ m/s$.

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