Question

A spherical drop of water carrying a charge of 25 pC has a potential of 670 V at its surface (with V = 0 at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?

Answer 1

Answer: a) 335.8 μm; b) keeping the same radius, the new has double potential, V=1340V so if teh radius is also double the potentail is the same (V=670V).

Explanation: In order to explain this problem we have to consider the potential given for sphere respect to infinity ( V=0) in the form:

V=k*Q/R the we have

R=k*Q/V= 9*10^9*25*10^-12/670=335.8 *10^-6 m

When two drop join to form a single drop (considering with the same radius) we have:

V=k*2Q/R

So the new V is double the original,

V=9*10^9*2*25*10^-12/335.8*10^-6=1340V

if the final single drop has a 2R of radius so

V=k*2Q/2R= 670 V

It has the same original potential.