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
σ_{b}
of bound charges on the surface of
the dielectric.

(b) Express
σ_{b}
in terms of the electric field just inside the dielectric. Let
ε be the dielectric constant of the dielectric.

(c) By using the equations obtained in (a) and (b), show that

σ_{b}
= -[1/(2π)][(ε - 1)/(
ε +1)][qh/(h^{2 }+ y^{2 }+ z^{2})^{3/2}].

(d) Calculate the electric field **E**' due to
σ_{b} 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** = -[(ε - 1)/(ε + 1)]
(q^{2}/4h^{2})**i**.

Solution:

- Concepts:

Polarization, polarization charge densities - Reasoning:

The field of the point charge polarizes the dielectric and induces a surface polarization charge density, σ_{b}=**P∙n**. - Details of the calculation:

(a) The field due to a surface charge density σ is 2πσ**n**(Gaussian units) just outside the surface.

Therefore**E**_{out}(0,y,z) = 2πσ_{b}(y,z)**i**+ q/(h^{2}+ y^{2}+ z^{2})^{3/2}[-h**i**+ y**j**+ z**k**].

Just inside the surface**E**_{in}(0,y,z) = -2πσ_{b}(y,z)**i**+ q/(h^{2}+ y^{2}+ z^{2})^{3/2}[-h**i**+ y**j**+ z**k**].

(b)**P**= [(ε - 1)/4π]**E**_{in}, σ_{b}=**P∙n**= [(ε - 1)/4π]E_{x_in}(0,y,z)

(c) σ_{b}= -[(ε - 1)/4π][2πσ_{b}+ qh/(h^{2}+ y^{2}+ z^{2})^{3/2}].

σ_{b}= -[1/(2π)][(ε - 1)/(ε + 1)][qh/(h^{2}+ y^{2}+ z^{2})^{3/2}].

(d)**E**'(h,0,0) =**i**∫hσ_{b}dA/(h^{2}+ y^{2}+ z^{2})^{3/2}

is the field due to σ_{b}at the position of q.

**E**'(h,0,0) = -**i**(q/(2π)) ((ε - 1)/(ε + 1)) ∫_{0}^{2π}dφ∫_{0}^{∞}h^{2}ρdρ/(h^{2}+ ρ^{2})^{3}

= -**i**(q/(2h)^{2}) (ε - 1)/(ε + 1).

This is the same field as that of a point charge q' = -q(ε - 1)/(ε + 1) located at (-h, 0, 0).

(e)**F**= q**E**' = -q^{2}(ε - 1)/[(ε + 1)4h^{2}]**i**.

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 ρ_{0}. They are floating with their
centers 1 km above a fresh-water mountain lake of
non-conducting pure water with dielectric constant
ε = 1.7ε_{0}. The bottom surface of each cloud is at
a potential of 10^{6} volts compared to the lake immediately below it.

(a) What is the charge density
ρ_{0}?

(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:**r**_{1}= (0, 0,1 )

Center of cloud 2:**r**_{2}= (2, 0, 1)

The potential in the region z > 0 can be found by postulating two image charges.

Center of image cloud 1:**r**_{1}' = (0, 0, -1)

Center of image cloud 2:**r**_{2}' = (2, 0, -1)

The charge density of each image cloud is

ρ' = -ρ_{0}(ε - ε_{0})/(ε + ε_{0}) = -(0.7/2.7)ρ_{0}= -0.26ρ_{0}.

Find the potential at**r**= (0, 0, 0):

Φ(**r**) = [ρ_{0}(4/3)π(0.5 km)^{3}/(4πε_{0})][1/(1 km) - 0.26/(1 km) + 1/(5^{½}km) - 0.26/(5^{½}km)]

= 4.47*10^{4}(ρ_{0}/ε_{0})m^{2}.

Find the potential at**r**= (0, 0, 0.5):

Φ(**r**) = [ρ_{0}(4/3)π(0.5 km)^{3}/(4πε_{0})][1/(0.5 km) - 0.26/(1.5 km) + 1/((4.25)^{½}km) - 0.26/((6.25)^{½}km)]

= 9.20*10^{4}(ρ_{0}/ε_{0})m^{2}.

ΔΦ = 10^{6}V = (9.20 - 4.47)10^{4}(ρ_{0}/ε_{0})m^{2}, ρ_{0}= 1.87*10^{-10}C/m^{3}.

(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:

F_{x}= [ρ_{0}V/(4πε_{0})][ρ_{0}V/(4 km^{2}) + (2/8^{½})ρ'V/(8 km^{2})]

F_{z}= [ρ_{0}V/(4πε_{0})][ρ'V/(4 km^{2}) + (2/8^{½})ρ_{0}V/(8 km^{2})]

Here V = (4/3)π(0.5 km)^{3}.

**F**= F_{x}**i**+ F_{z}**k**Force on cloud 2:

**F**= -F_{x}**i**+ F_{z}**k**from symmetry.

(c) n = c/v = (εμ/ε_{0}μ_{0})^{½}. Assume μ = μ_{0}, then n = (ε/ε_{0})^{½}= (1.7)^{½}= 1.3.