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Electrostatic Potential And Capacitance

Question

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n small drops of same size are charged to V volt each. They collapse to form a bigger drop. Calculate the capacity and potential of the bigger drop.

Solution

Let,
'r' represent the radius of small drop and,
'R' represent the radius of bigger drop.
Then,

\frac{4}{3} {πR}^{3} = n \times \frac{4}{3} {πr}^{3} \Rightarrow R = n^{1 / 3} r

This implies, the capacity of the bigger drop is n^1/3 times the capacity of each small drop.

Ptential of 1 big drop is given by V=n.

\frac{kq}{R}

= \frac{nq}{4 {πε}_{0} R} = \frac{nq}{4 {πε}_{0} n^{1 / 3} r} = n^{2 / 3} \frac{q}{4 {πε}_{0} r}

So, the potential of the bigger drop is n^2/3 times the potential of the smaller drop.

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