The potential energy of a system of two point | vedclass

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Question

(B) The potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by the formula: $U = \frac{1}{4\pi\

Vedclass · Accepted Answer

$3.6 	imes 10^{-2} \, J$

Vedclass · Answer

(B) The potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by the formula: $U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$.Given: $q_1 = q_2 = -2 \, \mu C = -2 	imes 10^{-6} \, C$,$r = 1 \, m$,and $k = \frac{1}{4\pi\epsilon_0} = 9 	imes 10^9 \, N \cdot m^2/C^2$.Substituting the values:$U = \frac{(9 	imes 10^9) 	imes (-2 	imes 10^{-6}) 	imes (-2 	imes 10^{-6})}{1}$$U = 9 	imes 10^9 	imes 4 	imes 10^{-12}$$U = 36 	imes 10^{-3} \, J = 3.6 	imes 10^{-2} \, J$.

The potential energy of a system of two point negative charges,each of magnitude $2 \, \mu C$,placed $1 \, m$ apart in air is:

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