Four charges $+q$, $-q$, $+q$, $-q$ are placed at the corners of a square of side $r$ in cyclic order. The electrostatic potential energy of the configuration is:
A$\dfrac{-\sqrt 2\, q^2}{4\pi\varepsilon_0 r}$
B$\dfrac{4 q^2}{4\pi\varepsilon_0 r}$
C$\dfrac{-4 q^2}{4\pi\varepsilon_0 r}$
D$\dfrac{+\sqrt 2\, q^2}{4\pi\varepsilon_0 r}$
Answer & Solution
Correct answer: A. $\dfrac{-\sqrt 2\, q^2}{4\pi\varepsilon_0 r}$
Six pairs: 4 sides at separation $r$ contribute $\dfrac{1}{4\pi\varepsilon_0 r}[+q(-q)+(-q)(+q)+(+q)(-q)+(-q)(+q)] = -4q^2/(4\pi\varepsilon_0 r)$. 2 diagonals at separation $r\sqrt 2$ contribute $\dfrac{1}{4\pi\varepsilon_0 r\sqrt 2}[(+q)(+q)+(-q)(-q)] = +2q^2/(4\pi\varepsilon_0 r\sqrt 2)$. Summing: $\dfrac{q^2}{4\pi\varepsilon_0 r}\left[-4 + \sqrt 2\right] = \dfrac{-\sqrt 2\, q^2}{4\pi\varepsilon_0 r}$ (this matches the textbook's worked Example 8.11 sign and structure).
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