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(C) 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 the formula: $U =

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will decrease

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(C) 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 the formula: $U = \frac{k q_1 q_2}{r}$.Here,the charges are $q_1 = +q$ and $q_2 = -q$. Therefore,the potential energy is $U = \frac{k(q)(-q)}{r} = -\frac{k q^2}{r}$.As the positive charge approaches the negative charge,the separation distance $r$ decreases.Since $r$ is in the denominator and the potential energy is negative,as $r$ decreases,the magnitude of the negative value increases,which means the value of $U$ becomes more negative.Therefore,the potential energy of the system decreases.

$A$ moving positive charge approaches a negative charge. What will happen to the potential energy of the system?

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