A spherical drop of water carrying a charge of `3 xx 10^(-19)C` has a potential of 500 V at its surface. What is the radius of the drop ? If two drops of the same charge and the same radius combine to form a single spherical drop, what is the potential at the surface of the new drop ?

Explanation

The potential V is given V` = ( q)/( 4 pi epsilon_(0)r)`
Here, `q = 3 xx 10^(-19)C` and `V = 500` volts
`:. R= 0.54 cm`
Volume of one drop is `(4)/( 3) pi r^(3)`
Total volume of both drop is `(8)/(3) pi r^(3)`
Let r' be the radius of the new drop formed , equating the volumes, we have `(4)/(3) pi r^('3) = (8)/( 3) pi r^(3)`
This gives `r' = ( 2^(1/3))r`
charge on the new drop `= 2 q = 6 xx 10^(-19) C`
New potential `V = (2 q)/( 4pi epsilon_(0) r^('))`
`= 794 ` volts.