When two electrons (e^-) move towards each ot | vedclass

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

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(B) The electrostatic potential energy $(U)$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$.For two electrons,$q_1 = -e$ and $q_2 = -e$,so $U = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$.Since both charges are negative,the product $q_1 q_2$ is positive,making $U$ positive.As the two electrons move towards each other,the distance $r$ between them decreases.Since $U \propto \frac{1}{r}$,as $r$ decreases,the potential energy $U$ increases.

When two electrons $(e^-)$ move towards each other,the electrostatic potential energy of the system will:

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