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(D) The work done $W$ 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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(D) The work done $W$ in moving a charge in an electrostatic field is equal to the change in the electrostatic potential energy of the system.$W = U_f - U_i$Initially,the charge $(-q)$ is at point $A$ and $(+Q)$ is at point $B$,separated by a distance $\ell$. The initial potential energy is $U_i = \frac{1}{4 \pi \varepsilon_0} \frac{(-q)(Q)}{\ell} = -\frac{1}{4 \pi \varepsilon_0} \frac{Qq}{\ell}$.Finally,the charge $(-q)$ is moved to point $C$,while $(+Q)$ remains at $B$. The distance between $C$ and $B$ is also $\ell$ because $ABC$ is an equilateral triangle.The final potential energy is $U_f = \frac{1}{4 \pi \varepsilon_0} \frac{(-q)(Q)}{\ell} = -\frac{1}{4 \pi \varepsilon_0} \frac{Qq}{\ell}$.Therefore,the net work done is $W = U_f - U_i = (-\frac{1}{4 \pi \varepsilon_0} \frac{Qq}{\ell}) - (-\frac{1}{4 \pi \varepsilon_0} \frac{Qq}{\ell}) = 0$.

$A$ charge $(-q)$ and another charge $(+Q)$ are kept at two points $A$ and $B$ respectively. Keeping the charge $(+Q)$ fixed at $B$,the charge $(-q)$ at $A$ is moved to another point $C$ such that $ABC$ forms an equilateral triangle of side $\ell$. The net work done in moving the charge $(-q)$ is

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