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(C) The potential energy of a charge $q$ at a point is given by $U = qV$,where $V$ is the electric potential due to the fixed charges.Initial position

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decrease by $\frac{4q^2}{3\pi\varepsilon_0 d}$

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(C) The potential energy of a charge $q$ at a point is given by $U = qV$,where $V$ is the electric potential due to the fixed charges.Initial position: $x_i = 0$. The distances from $4q$ and $-q$ are $r_1 = d/2$ and $r_2 = d/2$.Initial potential $V_i = \frac{1}{4\pi\varepsilon_0} \left( \frac{4q}{d/2} + \frac{-q}{d/2} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{3q}{d/2} \right) = \frac{6q}{4\pi\varepsilon_0 d}$.Final position: $x_f = d$. The distances from $4q$ and $-q$ are $r_1' = d + d/2 = 3d/2$ and $r_2' = d - d/2 = d/2$.Final potential $V_f = \frac{1}{4\pi\varepsilon_0} \left( \frac{4q}{3d/2} + \frac{-q}{d/2} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{8q}{3d} - \frac{2q}{d} \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{8q - 6q}{3d} \right) = \frac{2q}{12\pi\varepsilon_0 d} = \frac{q}{6\pi\varepsilon_0 d}$.Change in potential energy $\Delta U = q(V_f - V_i) = q \left( \frac{q}{6\pi\varepsilon_0 d} - \frac{6q}{4\pi\varepsilon_0 d} \right) = \frac{q^2}{4\pi\varepsilon_0 d} \left( \frac{2}{3} - 6 \right) = \frac{q^2}{4\pi\varepsilon_0 d} \left( -\frac{16}{3} \right) = -\frac{4q^2}{3\pi\varepsilon_0 d}$.The negative sign indicates a decrease in potential energy by $\frac{4q^2}{3\pi\varepsilon_0 d}$.

Two point charges $4q$ and $-q$ are fixed on the $x$-axis at $x = -d/2$ and $x = d/2$,respectively. If a third point charge $q$ is moved from the origin to $x = d$ along the semicircle as shown in the figure,the potential energy of the charge will:

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