A spherical drop of water bas `3 xx 10^(10)` C amount of charge residing on it. 500 V electric potential exists on its surface. Calculate the radius of this drop. If eight such drops (having identical charge and radii) combine to form a single drop, calculate the electric potential on the surface of the new drop. `(k = 9 xx 10^(9)` SI)

Explanation

Suppose R is the radius. The charge and potential on the surface are `3 xx 10^(10)` C and 500 V.
`:. V=(kQ)/(R)`
`:. R=(kQ)/(V)`
`= (9xx10^(9)xx3xx10^(-10))/(500)`
`= 5.4 xx10^(-3)` m
`= 0.54 xx10^(-2)` m
`:. R = 0.54 ` cm
The big drop is formed from combined 8 drops. The radius and volume of the big drop are R. and V..
`:. V = 8 V`
`:. (4)/(3) pi (R)^(3) = 8xx(4)/(3) pi (R)^(3)`
`:. (R)^(3) = 8(R)^(3) :. R = 2R`
`:. R . = 2xx0.54= 1.08 cm = 1.08 xx10^(-2)`m
The total charge on big drop
`Q=8xx3xx10^(-10)= 24 xx10^(10)` C
`:.` The electric potential of big drop
`V = (kQ)/(R)= (9xx10^(9)xx24xx10^(10))/(1.08xx10^(-2))`
`:. V = 2000 V `
Second Method :
Suppose radius of drop is R . Charge on its surface is `3xx10^(10)` C. Hence potential on its surface is 500 V.
`:. V = (kQ)/(R)`
`:. R = (kQ)/(V) = (9xx10^(9)xx3xx10^(-10))/(500)`
`:. R = (2.7)/(500) = 0.0054 ` m
`:. R = 0.54 ` cm
Potential on surface of new drop in given condition
`V.=n^(2/(3))V`
`= (8)^((2)/(3))xx500`
`=4xx500`
`:. V = 2000V`