Classical Electrodynamics · Problem 4.13

Magnetostatics, Faraday's Law, Quasi-Static Fields — John David Jackson

Physics — E&M · liquid-dielectric-susceptibility-capillary-rise

Problem

Two long, coaxial, cylindrical conducting surfaces of radii

a

and

b

are lowered vertically into a liquid dielectric. If the liquid rises an average height

h

between the electrodes when a potential difference

V

is established between them, show that the susceptibility of the liquid is

\chi_e = \frac{(b^2 - a^2)\rho g h \ln(b/a)}{\epsilon_0 V^2}

where

\rho

is the density of the liquid,

g

is the acceleration due to gravity, and the susceptibility of air is neglected.