Each of 27 identical mercury drops is charged to a potential of 10V. If the drops coalesce to form a big drop, what will be its potential? Calculate the ratio of the energy of the big drop to that of a small drop.

Explanation

Let the radius of each small drop be r.
Capacitance, `C_1 = 4pi in_0r`
Charge of each small drop, `q = C_1V_1= 10C_1`
Total charge, `Q= 27Q = 27xx 10C_1 = 270C_1`
If R be the radius of the big drop, then according to the question,
`(4)/(3)piR^3 = 27xx(4)/(3)pi r^3" or " R = 3r`
Capacitance of the bog drop, `C_2 = 4pi in_0*3r= 3C_1`
Potential of the big drop, `V_2=(Q)/(C_2)=(270C_1)/(3C_1)= 90V`
Energy of a small drop, `E_1=(1)/(2)C_1V_1^2= (1)/(2)xxC_1 xx(10)^2`
Energy of the big drop, `E_2=(1)/(2)C_1V_2^2= (1)/(2)xx3C_1xx(90)^2`
`:. (E_2)/(E_1)= (3C_1)/(C_1)xx((90)^2)/((10)^2)= (243)/(1)`
`E_2 : E_1= 243 : 1`.