A spherical drop of water carrying a charge of 43 pC has a potential of 540 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?

Explanation:

Given that,

Charge on a spherical drop of water is 43 pC

The potential at its surface is 540 V  

(a) The electric potential on the surface is given by :

r is the radius of the drop

(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,

Now the charge on the new drop is 2q. New potential is given by :

Hence, the radius of the drop is and the potential at the surface of the new drop is 856.79 V.

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explanation:

drying towels at the beach