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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{k q_1 q

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will increase

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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 $U = \frac{k q_1 q_2}{r}$,where $k$ is Coulomb's constant.For a proton $(+e)$ and an electron $(-e)$,the potential energy is $U = \frac{k(e)(-e)}{r} = -\frac{k e^2}{r}$.As the proton moves away from the electron,the separation distance $r$ increases.Since $r$ is in the denominator and the potential energy is negative,as $r$ increases,the value of $-\frac{k e^2}{r}$ becomes less negative (i.e.,it approaches zero from the negative side).Therefore,the potential energy of the system increases.

$A$ proton is moving away from an electron. Find the change in the potential energy of the system.

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