Question

1000 identical drops of mercury are charged to a potential of 1 volt each with joined to form a single drop the potential of each of this job will be

Answer 1

first of all, find radius of bigger drop.

Let radius of smaller drop is r

then, volume of bigger drop = 1000 × volume of smaller drop

or, 4/3 πR³ = 1000 × 4/3 πr³

or, R = 10r ......(1)

we know, charge = potential × capacitance

capacitance of a smaller drop = 4π$\epsilon_0$r

so, capacitance of 1000 smaller drops = 1000 × 4π$\epsilon_0$r

so, net charge = 1 × (1000 × $\epsilon_0$)

= 1000 × 4π$\epsilon_0$r

again, capacitance of bigger drop = 4π$\epsilon_0$R = 4π$\epsilon_0$(10r) [ from equation (1),

Let potential is V.

charge on bigger drop = 4π$\epsilon_0$ (10r) × V

so, charge on bigger drop = net charge on 1000 smaller drops

or, 4π$\epsilon_0$ (10r) × V = 1000 × 4π$\epsilon_0$r

or, V = 100 volts

hence, potential on bigger drop = 100 volts