1. Physics 505, Classical Electrodynamics
Homework 7
Due Thursday, 28th October 2004
Jacob Lewis Bourjaily
Problem 4.13
We are to consider two long, coaxial, cylindrical conducting surfaces of radii a; b (a < b) that are
lowered vertically into a liquid dielectric. The liquid rises an average height h between the two cylinders
when a potential di®erence V is established between them. Given that the density of the dielectric is
½0
1, we are to determine the susceptibility of the liquid.
First, let us assume that the cylinder is long so that the potential is independent of the µ and Á
coordinates. In this case, the Laplace equation for the space between the two cylinders is simply
1
½
@
@½
µ
½
@'
@½
¶
= 0;
and so we see that '(½) = ® log ½ + ¯ where ®; ¯ are determined by the boundary conditions. Because
we know that '(a) ¡ '(b) = V = ® log a + ¯ ¡ ® log b ¡ ¯ = ® log (a=b), we have that
® = V
log (a=b) :
We see that we can consistently set ¯ = 0 because we are only interested in potential di®erences.
Therefore,
'(½) = V
log (a=b)
log ½:
Now, using E = ¡r' and our general results about macroscopic electrostatics, we see that
E = V
log (b=a)
1
½
and D =
(
²V
log(b=a)
1
½ in the liquid
²0V
log(b=a)
1
½ above the liquid :
Therefore, following the discussion of electrostatic energy in dielectric media in Jackson's section 4.7,
we see that the energy density associated with the ¯eld con¯guration is given by
W =
1
2
Z
E ¢ Dd3x;
= ¼
Z b
a
d½ ½
"Z h
0
²
V 2
log2 (b=a)
1
½2 dz +
Z `
h
²0
V 2
log2 (b=a)
1
½2
#
;
= ¼V 2
log (b=a)
(h² + ²0` ¡ h²0) ;
where ` is the height of the cylinder. The term proportional to ` in the expression above for W can
be interpreted as the energy associated with the ¯eld con¯guration before lowing the cylinder into the
dielectric medium. Speci¯cally, we are interested in the energy required to create the new con¯guration
with the dielectric medium raised some height h above the other liquid (where we have set the level
of the dielectric outside of the cylinder to z = 0.). Therefore, the energy gained by lowering the two
cylinders is
¢W = ¼V 2h
log (b=a)
(² ¡ ²0) = ¼V 2h
log (b=a)
µ
²
²0
¡ 1
¶
²0 = ¼V 2hÂe²0
log (b=a) :
It is clear from elementary mechanics that this must be equal to the energy required to raise the mass
of the dielectric liquid in the gravitational ¯eld. The mass of the liquid is its density times its volume,
which is of course ½¼(b2 ¡ a2). And the energy required to raise this mass in a constant gravitational
¯eld g is simply ½0¼(b2 ¡ a2)gh.
Therefore, because the energy gained from the ¯eld con¯guration of the cylinders with the dielectric
medium must be the same as that required to raise the liquid, we have that
½0¼
¡
b2 ¡ a2¢
gh = ¼V 2hÂe²0
log (b=a) :
Therefore, it is clear that
) Âe =
½0gh
¡
b2 ¡ a2
¢
log (b=a)
²0V 2 :
`¶o¼²½ '¶²±²¶ ±²¶»®¶
1The subscript on ½0 is used to distinguish it from the radial coordinate ½.
1