What is the capacitance of a drop that results when two mercury spheres, each of radius R = 2.00mm merge?
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Assuming conservation of
volume, we find the radius of the combined spheres, then use C=4πε0R to find the capacitance. When the drops combine, the
volume is doubled. It is then V=2(4π/3)R3. The new radius R′ is
given by : 34π(R′)3=234πR3⇒R′=21/3R The new capacitance is : C′=4πε0R′=4πε021/3R=5.04πε0R With R=2.00mm, we obtain C=5.04π(8.85×10−12F/m)(2.00×10−3m)=2.80×10−13F.
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