# Dielectrics and Infinite Electric Cylinders

1. The space between the spaces of a parallel plate capacitor is filled with two slabs of linear dielectric material. The slabs have different dielectric constants but the same length L, width W and thickness d. (note the area of the top (or bottom) of the capacitor is 2*L*W). Slab 1 has a dielectric constant of e1=2 and slab 2 has a dielectric constant of e2=1.5. The voltage on the top plate is Vo and the bottom plate is grounded.

a) What is the electric Field E in each slab?

b) What is the Dielectric displacement D in each slab?

c) What is the polarization P in each slab?

d) What is the magnitude of Sf, the free surface charge adjacent to each slab?

e) What is the magnitude of Sb, the bound surface charge density in each slab?

f) What is the magnitude of Vb, bound volume charge density in each slab?

g) What is the capacitance of the system and how does it compare to the capacitance of the system with no dielectrics?

See the attachment for an illustration.

2) We have an infinite electric cylinder of radius R that has a polarization

Pbar=kr^2 rhat (a vector)

a) What is the bound volume density in the body of the cylinder

b) What is the bound surface density at r=R

c) Use Gauss' Law to find the electric field inside and outside the cylinder

d) Use a releationship between

E, D, and P to find the electric fields without using Gauss' Law.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attachments for the solution.

The space between the spaces of a parallel plate capacitor is filled with two slabs of linear dielectric material. The slabs have different dielectric constants but the same length L, width W and thickness d. (note the area of the top (or bottom) of the capacitor is 2*L*W). Slab I has a dielectric constant of and slab II has a dielectric constant . The voltage on the top plate is V0 and the bottom plate is grounded.

a) What is the electric Field E in each Slab

b) What is the Dielectric displacement D in each slab

c) What is the polarization P in each slab

d) What is the magnitude of , the free surface charge adjacent to each slab

e) What is the magnitude of , the bound surface charge density in each slab

f) What is the magnitude of , bound volume charge density in each slab

g) What is the capacitance of the system and how does it compare to the capacitance of the system with no dielectrics.

The potential difference across each slab is . We assume that , so the electric field inside the capacitor is constant.

Thus we get an electric field across the dielectrics (both):

The displacement field in slab I is:

And in slab II the displacement field is

The polarization in slab I is:

The polarization in slab II is:

Using Gauss' law at the top plate we have:

Thus (note that ):

Therefore above dielectric I the free charge density is:

And above dielectric II the free charge is:

The bound surface charge density in slab I is (there ):

And in slab II the bound surface charge density is:

Since in both slabs the polarization is constant (no dependence on coordinates) there is no bound volume charge density:

The capacitance is given as:

We know that the total bound charge is zero (the surface charge on the top of the slab is negative the bound surface charge at the bottom).

So we are left with a total free charge of

For an air-filled parallel plate capacitor of plate area of 2A, the capacitance is:

Hence:

Note that this configuration is actually two individual capacitors each with plate area A, one filled with dielectric and the other filled with dielectric and these two capacitors are connected in parallel - that is, their capacitance is additive.

We have an infinite electric cylinder of radius R that has a polarization

a) What is the bound volume density in the body of the cylinder

b) What is the bound surface density at r=R

c) Use Gauss' Law to find the electric field inside and outside the cylinder

d) Use a relationship between E, D, and P to find the electric fields without using Gauss' Law.

The polarization is:

Thus, the bound surface charge density is:

And the bound volume charge density is (using cylindrical coordinates):

Gauss law states that:

Our Gauss surface is a concentric cylinder of length L and radius r.

The electric field on the envelope of the cylinder is radial and its magnitude constant.

Outside the cylinder the total enclosed charge is:

Thus, there is no electric field outside the cylinder.

Inside the cylinder we have no contribution from the surface charge and we are left with:

So the electric field is:

Since there are no free charges, the displacement field everywhere is 0 by definition.

Thus:

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