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Question

(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 $U = \frac{K q_1 q

Vedclass · Accepted Answer

$\frac{K q_1 q_2 x}{d(d-x)}$

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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 $U = \frac{K q_1 q_2}{r}$.Initial potential energy $U_i$ at distance $d$ is $U_i = \frac{K q_1 q_2}{d}$.When charge $q_2$ is moved towards $q_1$ by a distance $x$, the new separation becomes $r' = d - x$.The final potential energy $U_f$ is $U_f = \frac{K q_1 q_2}{d - x}$.The increase in potential energy $\Delta U$ is given by $\Delta U = U_f - U_i$.Substituting the values:$\Delta U = \frac{K q_1 q_2}{d - x} - \frac{K q_1 q_2}{d}$$\Delta U = K q_1 q_2 \left( \frac{1}{d - x} - \frac{1}{d} \right)$$\Delta U = K q_1 q_2 \left( \frac{d - (d - x)}{d(d - x)} \right)$$\Delta U = \frac{K q_1 q_2 x}{d(d - x)}$.

Two point charges $q_1$ and $q_2$ are separated by a distance $d$. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ by a distance $x$? $(x < d)$ (where $\frac{1}{4 \pi \varepsilon_0} = K$ is a constant).

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