A conducting cylinder immersed in a dielectric liquid
A conducting cylinder of radius a and height L ≫ a is immersed for a
depth L − h (with h ≫ a) in a dielectric liquid having permittivity ε. A
free charge Q is located on the cylinder. Boundary effects are negligible.
Find
a) the electric field Ea at the surface of the cylinder and the surface charge
density;
b) the electric field in the region between the surface of the cylinder and
the container of the liquid (a < r < b);
c) whether the electrostatic force on the cylinder act either to dunk it
into the liquid or to emerge.
d) How is an equilibrium position for the cylinder determined, taking the
Archimedes’ principle into account? (Let M be the mass of the cylinder
and ρ the density of the liquid).
1
b
a
Q
h
ε
L−h
Solution
a) The electric field is normal to the surface of the cylinder and it is continous at the interface
between the dielectric liquid and the vacuum being parallel to the interface itself. Let σ1 and σ2 be
the surface density of free charge on the region of the conductor facing the vacuum and the dielectric,
respectively. The following relations hold,
ε 0 E a = σ1 ,
ε r ε 0 E a = σ2 ,
σ1 h + σ2 (L − h) = Q/(2πa) ,
(1)
from which we obtain
Ea ==
Q
Q
=
.
2πε0 a[h + εr (L − h)]
2πε0 a[(εr − 1)h + εr L]
(2)
b) By applying Gauss’ theorem to a cylindrical surface, coaxial with the conductor and having radius
r such that b > r > a we obtain
2πrE(r) = Qtot ,
(3)
where Qtot is the total charge, which includes both free and polarization charges. For r → a we have
2πaEa = Qtot , thus we can write
E(r) = Ea a/r .
c) The electrostatic energy of the system is
Z
Z
ε(z) 2
1
ε0
2
3
Ues =
E (r)2πrdr [h + εr (L − h)] ∝
E (r)d x =
.
2
2
[h + εr (L − h)]
(4)
(5)
The force in vertical direction is given by minus the derivative of Ues with respect to h:
F =−
εr − 1
dUes
∝−
<0,
dh
[h + εr (L − h)]
(6)
which is negative. This means that electrostatic forces tend to decrease h, i.e. to dunk the cylinder
into the liquid.
d) The variation of gravitational energy for an infinitesimal displacement dh is
dUg = M gdh − ρg(L − h)πa2 dh .
The equilibrium position is obtained by posing dUg + dUes = 0 and solving for h.
2
(7)