A charge of 10, e.s.u. is placed at a distanc | vedclass

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(D) The potential energy $U$ of a system of point charges in the $CGS$ system (where $k = 1$) is given by $U = \sum \frac{q_i q_j}{r_{ij}}$.Here,the c

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$250$

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(D) The potential energy $U$ of a system of point charges in the $CGS$ system (where $k = 1$) is given by $U = \sum \frac{q_i q_j}{r_{ij}}$.Here,the charge $q_1 = 10\, e.s.u.$ is interacting with $q_2 = 40\, e.s.u.$ at a distance $r_1 = 2\, cm$ and $q_3 = 20\, e.s.u.$ at a distance $r_2 = 4\, cm$.The total potential energy of the $10\, e.s.u.$ charge is the sum of the interaction energies with the other two charges:$U = \frac{q_1 q_2}{r_1} + \frac{q_1 q_3}{r_2}$$U = \frac{10 	imes 40}{2} + \frac{10 	imes 20}{4}$$U = \frac{400}{2} + \frac{200}{4}$$U = 200 + 50 = 250\, ergs$.

$A$ charge of $10\, e.s.u.$ is placed at a distance of $2\, cm$ from a charge of $40\, e.s.u.$ and $4\, cm$ from another charge of $20\, e.s.u.$ The potential energy of the charge $10\, e.s.u.$ is (in $ergs$)

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