Question

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Assuming conservation of
volume, we find the radius of the combined spheres, then use $C=4πε_{0}R$ to find the capacitance. When the drops combine, the
volume is doubled. It is then $V=2(4π/3)R_{3}$. The new radius $R_{′}$ is
given by :

$34π (R_{′})_{3}=234π R_{3}⇒R_{′}=2_{1/3}R$

The new capacitance is :

$C_{′}=4πε_{0}R_{′}=4πε_{0}2_{1/3}R=5.04πε_{0}R$

With $R=2.00mm,$ we obtain $C=5.04π(8.85×10_{−12}F/m)(2.00×10_{−3}m)=2.80×10_{−13}F.$

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