Question

Problem 2 A spherical drop of water carrying a charge of 29 pC has a potential of 660 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

(a)
The potential on the surface of the drop is
V = kq / r
Here k = 9*10^9 N.m2 / C2 , q = 29 pc= 29*10^-12, v = 660 volt
Rearranging and substitute all values in the above equation , we get
The radius of the sphere is r = kq / V
= ( 9*10^9 N.m2 / C2 ) ( 29*10^-12 C ) / ( 660 V)
= 0. 395 *10^-3 m
= ( 0. 395 *10^-3 m ) ( 1 mm / 10^-3 m )
= 0.395 mm
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(b)
The total charge on the new spherical drop is 2q = 2 ( 29*10^-12 C )

= 58*10^-12 C
The volume of the new sphere is sum of the volumes of the two spherical drops is
V = ( 4/3) π r^3 + ( 4/3) π r^3
( 4/3) π R^3 = 2 ( 4/3) π r^3
Solving the above equation, we get
The radius of the new spherical drop is
R = cube root of ( 2r^3 )
  
R = cube root of ( 2( 0.395 *10^-3 )^3 )

= 0.497 *10^-3 m
The potential on the new spherical drop is
V = k (2q ) / R
= ( 9*10^9 ) ( 58*10^-12 C ) / (0.497 *10^-3 m )
= 1050.3 V