A conducting sphere between two dielectrics
A conducting sphere of radius R floats over a liquid with dielectric permittivity ε1 , with the level of the liquid corresponding to the equator of the sphere as shown in the figure. The
gaseous medium above the liquid has permittivity ε2 < ε1 . A
free charge Q is located on the conducting sphere. Find
a) the electric field at the surface of the sphere,
b) the surface densities of free and polarization charges,
c) the sign of the electrostatic force on the sphere.
1
ε2
ε1
Q
Solution
a) The electric field is normal to the surface of the sphere.
Since ∇ × E = 0 we can assume that the field is of uniform amplitude over each half-surface of the
sphere, and also that the field is continuous across the dielectric interface being parallel to it. The
flux of D = εE across a closed surface enclosing the sphere is equal to the free charge Q. Thus,
taking a spherical surface concentric with the sphere and of arbitrarily close radius we can write
Q = Φ(εE) = 2πR2 (ε1 E1 + ε2 E2 ) = 2πR2 (ε1 + ε2 )E ,
(1)
from which we obtain
E=
Q
1
.
2
2πR ε1 + ε2
(2)
b) At the surface of the sphere the free charge density σf = (εE)⊥ , thus we obtain
σf,i =
εi
Q
,
2πR2 ε1 + ε2
i = 1, 2 .
(3)
The surface density of polarization charges is σp = P⊥ = ε0 χE⊥ with χ = 1 − ε/ε0 where P is the
polarization density, and we obtain
σp,i =
Q ε0 − εi
,
2πR2 ε1 + ε2
i = 1, 2 .
(4)
c) The electrostatic tension P (not to be confused with the polarization density!) on each half-surface
of the conducting sphere is
1
1 2
= ε2i E
Pi = σf,i
2
2
i = 1, 2 ,
(5)
thus P2 > P1 because ε2 > ε1 holds: the total force drags the sphere towards the medium of higher
permittivity.
2