Two identical metal balls with charges + 2 Q and -Q are separated by some distance and exert a force F on each other. They are joined by a conducting wire, which is than removed. The force between them will now be

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F/2

F/4

F/8

answer is D.

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Detailed Solution

F=14πε0×2Q×Qr2=14πε0×2Q2r2After joining by a conducting wire, the charge on each metal ball will be(2Q−Q)/2=Q/2∴ F′=14πε0×(Q/2)(Q/2)r2 =14πε0Q24r2=F8

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