A spherical drop of water carrying a charge of 38 pC has a potential of 710 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?

Question

Hannah Odom

Physics

24 April, 02:25

A spherical drop of water carrying a charge of 38 pC has a potential of 710 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?

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Scott Ferguson · Answer 1

Answer: a) r = 4.82 * 10^-4 m; b) 1420 V

Explanation: In order to solve this problem we have to take into account that potential for a sphere respec to V=0 at the infinity, which is given by:

V=k*Q/r where r is the radius of the drop

then we have

r=k*Q/V=9*10^9*38pC/710V = 4.82 * 10^-4 m

Finally if we join two drop to form one with the same radius but with twice charge the resultant potential is:

V = k*2*Q / (r) = 710*2 = 1420 V