Section E

The closed surface integral sums the outward electric flux:

E

is the electric field,

d S

its outward vector area,

Q

the total enclosed charge and

ϵ_{0}

the vacuum permittivity. Gauss's law relates that flux to enclosed charge for any closed surface. High symmetry can make the field's magnitude constant on a suitable Gaussian surface and fix its direction; without symmetry, flux alone does not determine the local field.

For an isolated conducting sphere, charges reside at its surface and the exterior field is radial. Thus

E (r) = \frac{Q}{4 π ϵ _{0} r ^{2}} r (r > a), V (a) = \int_{a}^{\infty} E (r) d r = \frac{Q}{4 π ϵ _{0} a}, C = 4 π ϵ_{0} a .

(1)

The conductor has zero interior field. Integrating exterior field energy gives

U = \int_{a}^{\infty} \frac{1}{2} ϵ_{0} (\frac{Q}{4 π ϵ _{0} r ^{2}})^{2} 4 π r^{2} d r = \frac{Q ^{2}}{8 π ϵ _{0} a} .

(2)

Let the initial charge per drop be

q = 4 π ϵ_{0} a V_{1}

. After disconnecting the charging supply, charge is conserved. Volume conservation gives final radius

A = 2^{1/3} a

, total charge

2 q

and

V_{f} = \frac{2 q}{4 π ϵ _{0} A} = 2^{2/3} V_{1} .

(3)

Initial mutual energy is negligible at the stipulated large separation. Hence

U_{i} = \frac{q ^{2}}{4 π ϵ _{0} a}, U_{f} = \frac{4 q ^{2}}{8 π ϵ _{0} 2 ^{1/3} a} = 2^{2/3} U_{i}, U_{i} / U_{f} = 2^{- 2/3} .

(4)

The electrostatic energy increase on equal-drop coalescence is positive. It can be supplied by the decrease in surface energy during coalescence and/or by external mechanical work bringing like charges together. Remaining energy can become motion and heat. If the charge is too large, surface-energy release alone need not suffice to make merging spontaneous; conservation does not assert that it will. Keeping the drops connected to a voltage supply would instead be a different charge-balance problem.