Question:
Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, -2) and (0, -2). The work required to put a fifth charge Q at the origin of the coordinate system will be -
Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, -2) and (0, -2). The work required to put a fifth charge Q at the origin of the coordinate system will be -
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Work needed = Q times the potential at the origin due to the four charges.
Updated On: Nov 5, 2026
- \(\frac{Q^2}{4\pi\varepsilon_0}\)
- \(\frac{Q^2}{2\sqrt2\,\pi\varepsilon_0}\)
- \(\frac{Q^2}{4\pi\varepsilon_0}\left(1+\frac{1}{\sqrt3}\right)\)
- \(\frac{Q^2}{4\pi\varepsilon_0}\left(1+\frac{1}{\sqrt5}\right)\)
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The Correct Option is D
Solution and Explanation
Distances of the four charges from the origin: \((0,2)\): 2; \((0,-2)\): 2; \((4,2)\): \(\sqrt{16+4}=2\sqrt5\); \((4,-2)\): \(2\sqrt5\).
Potential at origin: \(V=\frac{Q}{4\pi\varepsilon_0}\left[\frac12+\frac12+\frac{1}{2\sqrt5}+\frac{1}{2\sqrt5}\right]=\frac{Q}{4\pi\varepsilon_0}\left(1+\frac{1}{\sqrt5}\right)\).
Work \(W=QV=\frac{Q^2}{4\pi\varepsilon_0}\left(1+\frac{1}{\sqrt5}\right)\).
Answer: (4) \(\frac{Q^2}{4\pi\varepsilon_0}\left(1+\frac{1}{\sqrt5}\right)\)
Potential at origin: \(V=\frac{Q}{4\pi\varepsilon_0}\left[\frac12+\frac12+\frac{1}{2\sqrt5}+\frac{1}{2\sqrt5}\right]=\frac{Q}{4\pi\varepsilon_0}\left(1+\frac{1}{\sqrt5}\right)\).
Work \(W=QV=\frac{Q^2}{4\pi\varepsilon_0}\left(1+\frac{1}{\sqrt5}\right)\).
Answer: (4) \(\frac{Q^2}{4\pi\varepsilon_0}\left(1+\frac{1}{\sqrt5}\right)\)
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