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(C) Initial potential energy of the system of three charges at the vertices of an equilateral triangle with side $r = 0.5 \ m$ and $q = 2 	imes 10^{-

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$72$

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(C) Initial potential energy of the system of three charges at the vertices of an equilateral triangle with side $r = 0.5 \ m$ and $q = 2 	imes 10^{-3} \ C$ is:$U = 3 	imes \frac{k q^2}{r} = 3 	imes \frac{9 	imes 10^9 	imes (2 	imes 10^{-3})^2}{0.5} = 3 	imes \frac{9 	imes 10^9 	imes 4 	imes 10^{-6}}{0.5} = 216 	imes 10^3 \ J$When one charge is moved to the midpoint of the line joining the other two charges,the new distances are: two charges are at distance $0.25 \ m$ from the moved charge,and the two stationary charges are at distance $0.5 \ m$ from each other.The final potential energy $U'$ is:$U' = \frac{k q^2}{0.25} + \frac{k q^2}{0.25} + \frac{k q^2}{0.5} = k q^2 \left( 4 + 4 + 2 ight) = 10 k q^2 = 10 	imes 9 	imes 10^9 	imes 4 	imes 10^{-6} = 360 	imes 10^3 \ J$The work done (energy required) is $\Delta U = U' - U = (360 - 216) 	imes 10^3 = 144 	imes 10^3 \ J$Given power $P = 2 \ kW = 2000 \ W$,the time taken $t$ is:$t = \frac{\Delta U}{P} = \frac{144 	imes 10^3}{2 	imes 10^3} = 72 \ s$

Three point charges of $2 \ mC$ each are kept at the vertices of an equilateral triangle of side $50 \ cm$. If the system is supplied energy at the rate of $2 \ kW$,the time taken to move one of the charges to the midpoint of the line joining the other two charges is (in $s$)

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