As shown in the figure below, a charge +2 tex | vedclass

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(C) The work done in moving a charge in an electrostatic field is equal to the change in the electrostatic potential energy of the system.$W = U_{f} -

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(C) The work done in moving a charge in an electrostatic field is equal to the change in the electrostatic potential energy of the system.$W = U_{f} - U_{i}$$U = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}$Initial potential energy $U_{i}$ when the charge $+5 	ext{ C}$ is at $A(2, 0) 	ext{ m}$:$U_{i} = \frac{1}{4 \pi \varepsilon_{0}} \frac{(2)(5)}{2} = \frac{1}{4 \pi \varepsilon_{0}} (5)$Final potential energy $U_{f}$ when the charge $+5 	ext{ C}$ is at $B(0, 2) 	ext{ m}$:$U_{f} = \frac{1}{4 \pi \varepsilon_{0}} \frac{(2)(5)}{2} = \frac{1}{4 \pi \varepsilon_{0}} (5)$Since the distance $r$ from the origin $O$ to point $A$ is $2 	ext{ m}$ and from the origin $O$ to point $B$ is also $2 	ext{ m}$, the potential energy remains the same.$W = U_{f} - U_{i} = 0 	ext{ J}$.

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