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(D) The potential energy $U$ of a system of point charges is given by $U = \sum \frac{1}{4\pi\epsilon_0} \frac{q_i q_j}{r_{ij}}$.Here, we have three c

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$-\frac{4q^2}{4\pi\epsilon_0x}$

Vedclass · Answer

(D) The potential energy $U$ of a system of point charges is given by $U = \sum \frac{1}{4\pi\epsilon_0} \frac{q_i q_j}{r_{ij}}$.Here, we have three charges:$q_A = q$ at point $A$.$q_B = q$ at point $B$.$q_C = -3q$ at point $C$.Given distances:$AC = x$.$AB = 3x$, so $CB = AB - AC = 3x - x = 2x$.Now, calculate the potential energy for each pair:$U_{AB} = \frac{1}{4\pi\epsilon_0} \frac{q_A q_B}{AB} = \frac{1}{4\pi\epsilon_0} \frac{q \cdot q}{3x} = \frac{q^2}{12\pi\epsilon_0x}$.$U_{AC} = \frac{1}{4\pi\epsilon_0} \frac{q_A q_C}{AC} = \frac{1}{4\pi\epsilon_0} \frac{q \cdot (-3q)}{x} = -\frac{3q^2}{4\pi\epsilon_0x}$.$U_{CB} = \frac{1}{4\pi\epsilon_0} \frac{q_C q_B}{CB} = \frac{1}{4\pi\epsilon_0} \frac{(-3q) \cdot q}{2x} = -\frac{3q^2}{8\pi\epsilon_0x}$.Total potential energy $U = U_{AB} + U_{AC} + U_{CB} = \frac{1}{4\pi\epsilon_0x} [\frac{q^2}{3} - 3q^2 - 1.5q^2] = \frac{1}{4\pi\epsilon_0x} [\frac{q^2}{3} - 4.5q^2] = \frac{1}{4\pi\epsilon_0x} [\frac{q^2 - 13.5q^2}{3}] = -\frac{12.5q^2}{12\pi\epsilon_0x} \approx -\frac{4q^2}{4\pi\epsilon_0x}$ (approximately).Thus, the correct option is $D$.

Two equal positive charges each of value $q$ are placed at points $A$ and $B$, where $AB = 3x$. A third charge $-3q$ is placed at point $C$ at a distance $x$ from $A$ on the line segment $AB$. The potential energy of the system is nearly ($\epsilon_0$ = permittivity of free space).

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