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(C) The potential energy $U$ of a system of point charges is given by $U = k \sum \frac{q_i q_j}{r_{ij}}$.Initial potential energy $U_i$ with side $r_

Vedclass · Accepted Answer

$99 	imes 10^9 	ext{ J}$

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(C) The potential energy $U$ of a system of point charges is given by $U = k \sum \frac{q_i q_j}{r_{ij}}$.Initial potential energy $U_i$ with side $r_1 = 1 	ext{ m}$:$U_i = k \left[ \frac{1 	imes 2}{1} + \frac{2 	imes 3}{1} + \frac{3 	imes 1}{1} ight] = k [2 + 6 + 3] = 11k$.Final potential energy $U_f$ with side $r_2 = 0.5 	ext{ m}$:$U_f = k \left[ \frac{1 	imes 2}{0.5} + \frac{2 	imes 3}{0.5} + \frac{3 	imes 1}{0.5} ight] = k [4 + 12 + 6] = 22k$.The work done $W$ is the change in potential energy:$W = U_f - U_i = 22k - 11k = 11k$.Given $k = 9 	imes 10^9 	ext{ N m}^2 	ext{ C}^{-2}$,$W = 11 	imes 9 	imes 10^9 = 99 	imes 10^9 	ext{ J}$.

Three point charges $1 \text{ C}, 2 \text{ C}$ and $3 \text{ C}$ are placed at the corners of an equilateral triangle of side $1 \text{ m}$. The work done to move these charges to the corners of another equilateral triangle of side $0.5 \text{ m}$ is

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