A charge of 10 , text{esu} is placed at a dis | vedclass

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(C) The potential energy $U$ of a charge $q$ in an electric field $V$ is given by $U = qV$.In the $CGS$ system, the electrostatic constant $k = 1$.The

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$150$

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(C) The potential energy $U$ of a charge $q$ in an electric field $V$ is given by $U = qV$.In the $CGS$ system, the electrostatic constant $k = 1$.The potential $V_A$ at the position of the $10 \, 	ext{esu}$ charge due to the other two charges is:$V_A = \frac{q_B}{r_1} + \frac{q_C}{r_2}$Given $q_B = 40 \, 	ext{esu}$, $r_1 = 2 \, 	ext{cm}$, $q_C = -20 \, 	ext{esu}$, and $r_2 = 4 \, 	ext{cm}$.$V_A = \frac{40}{2} + \frac{-20}{4} = 20 - 5 = 15 \, 	ext{statvolts}$.The potential energy $U_A$ of the $10 \, 	ext{esu}$ charge is:$U_A = q_A 	imes V_A = 10 	imes 15 = 150 \, 	ext{ergs}$.

$A$ charge of $10 \, \text{esu}$ is placed at a distance of $2 \, \text{cm}$ from a charge of $40 \, \text{esu}$ and at a distance of $4 \, \text{cm}$ from another charge of $-20 \, \text{esu}$. The potential energy of the $10 \, \text{esu}$ charge in ergs is:

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