### The method of images with dielectrics

#### Problem:

A charge q is situated at the point x = h > 0, y = z = 0 outside a homogeneous dielectric which fills the region x < 0.
(a)  Write the electric fields just outside and just inside the dielectric in terms of the charge q and surface charge density σb of bound charges on the surface of the dielectric.
(b)  Express σb in terms of the electric field just inside the dielectric.  Let K be the dielectric constant of the dielectric.
(c)  By using the equations obtained in (a) and (b), show that
σb = -[1/(2π)][(K - 1)/(K + 1)][qh/(h2 + y2 + z2)3/2].
(d)  Calculate the electric field E' due to σb at the position (h,0,0) of the charge q.  Show that it can be interpreted as the field of an image charge q' situated at the point (-h,0,0).
(e) Show that the charge q experiences the force F = -[(K - 1)/(K + 1)] (q2/4h2)i.

Solution:

• Concepts:
Polarization, polarization charge densities
• Reasoning:
The field of the point charge polarizes the dielectric and induces a  surface polarization charge density, σb = P∙n.
• Details of the calculation:
(a)  The field due to a surface charge density σ is σ/(2ε0) n just outside the surface.
Therefore Eout(0,y,z) = 2πkσb(y,z) i + kq/(h2 + y2 + z2)3/2 [-hi + yj + zk].
Here k = 1/(4πε0).
Just inside the surface Ein(0,y,z) = -2πkσb(y,z) i + kq/(h2 + y2 + z2)3/2 [-hi + yj + zk].

(b) P = ε0(K - 1)]Ein,  σb = P∙n = ε0(K - 1)Ex_in(0,y,z)

(c)  σb = -ε0(K - 1)[2πkσb + kqh/(h2 + y2 + z2)3/2].
σb + ε0(K - 1)2πkσb = -ε0(K - 1)kqh/(h2 + y2 + z2)3/2.
σb = -[1/(2π)][(K - 1)/(K + 1)][qh/(h2 + y2 + z2)3/2].

(d)  E'(h,0,0) = ik∫hσbdA/(h2 + y2 + z2)3/2
is the field due to σb at the position of q.  The components perpendicular to i cancel.
E'(h,0,0) = -ik[q/(2π)][(K - 1)/(K + 1)]h20dφ∫0ρdρ/(h2 + ρ2)3
E'(h,0,0) = -ikq[1/(2π)][(K - 1)/(K + 1)]*(π/2) = -ikq[(K - 1)/(K + 1)]/(4h2).
This is the same field as that of a point charge q' = -q(K - 1)/(K + 1) located at (-h, 0, 0).

(e) F = qE' = -i kq2[(K - 1)/(K + 1)]/(4h2).

#### Problem:

Two spherical clouds, each of 1 km diameter, are separated by a distance of 2 km (center-to-center).  Each cloud is uniformly charged with charge density ρ0.  They are floating with their centers 1 km above a fresh-water mountain lake of non-conducting pure water with dielectric constant ε = 1.7ε0.  The bottom surface of each cloud is at a potential of 106 volts compared to the lake immediately below it.
(a)  What is the charge density ρ0?
(b)  What is the electrostatic force on each cloud?
(c)  What is the index of refraction of the lake water?

Solution:

• Concepts:
The method of images, Coulomb's law
• Reasoning:
The electrostatic force on each cloud is the same as that due to the charges on the other cloud and due to the images charges behind the dielectric-dielectric interface.
• Details of the calculation:

(a)  Pick a coordinate system:
Center of cloud 1: r1 = (0, 0,1 )
Center of cloud 2: r2 = (2, 0, 1)
The potential in the region z > 0 can be found by postulating two image charges.
Center of image cloud 1: r1' = (0, 0, -1)
Center of image cloud 2: r2' = (2, 0, -1)
The charge density of each image cloud is
ρ' = -ρ0(ε - ε0)/(ε + ε0) = -(0.7/2.7)ρ0  = -0.26ρ0.
Find the potential at r = (0, 0, 0):
Φ(r) = [ρ0(4/3)π(0.5 km)3/(4πε0)][1/(1 km) - 0.26/(1 km) + 1/(5½ km) - 0.26/(5½ km)]
= 4.47*10400)m2.
Find the potential at r = (0, 0, 0.5):
Φ(r) = [ρ0(4/3)π(0.5 km)3/(4πε0)][1/(0.5 km) - 0.26/(1.5 km) + 1/((4.25)½ km) - 0.26/((6.25)½ km)]
= 9.20*10400)m2.
ΔΦ = 106 V = (9.20 - 4.47)10400)m2,  ρ0 = 1.87*10-10 C/m3.
(b)  Take cloud 1.  The force on cloud 1 is the same as that due to cloud 2 and image clouds 1 and 2.
The force on cloud 1 due to each of these clouds is the same as that due to point charges with the same total charge at the center of the clouds.  It can be calculated assuming the total charge of cloud 1 is at its center.
Force on cloud 1:
Fx = [ρ0V/(4πε0)][ρ0V/(4 km2) + (2/8½)ρ'V/(8 km2)]
Fz = [ρ0V/(4πε0)][ρ'V/(4 km2) + (2/8½0V/(8 km2)]
Here V = (4/3)π(0.5 km)3.
F = Fxi + Fzk
Force on cloud 2:
F = -Fxi + Fzk  from symmetry.
(c)  n = c/v = (εμ/ε0μ0)½.  Assume μ = μ0, then n = (ε/ε0)½ = (1.7)½ = 1.3.