What is the Electric Field and Displacement in a Polarized Cylinder?

[stunner5000pt]

Homework Statement

Suppose we have a solid cylinder of radius a and length L and carries a uniform polarization parallel to its axis. Find the electric field everywhere.

Suppose we have a cylinderical cavity inside a large dielectric material which has electric displacement D0 and electric field E0 and polarziation P such taht
\vec{D} = \epsilon_{0} \vec{E_{0}} + \vec{P}. Then find the electric field and electric displacement within the cyldrical cavity

Homework Equations

Bound surface charge
\sigma_{b} = \vec{P}\cdot \hat{n}

Bound volume charge
\rho_{b} = - \vec{\nabla} \cdot \vec{P}

The Attempt at a Solution

uniform polairztaion so there is no bound volume charge
Since the polarization is parallel to the axis there is no surface charge on teh curved part of the cylinder. However there is a bound charge on the caps on the cylinder.

If the cylinder is aligned parallel to the Z axis and the polaraation pointsi nteh Z axis then the lower part of the top cap is -P

so now we have two circular caps which are parallel with opposite charge densities. SO the electric field is just like that of a parallel plates each wit ha cahrge density of P

sp the electric field inside if E = \frac{P}{\epsilon_{0}} (-\hat{z})

The electric field outside... Well there is no free charge so the electric displacement is zero
D = \epsilon_{0} E + P
0 = \epsilon_{0} E + P
E= \frac{P}{\epsilon_{0}} (\hat{-z})

is this correct??

Now for the part where this cylinder is the cavity. We assume that the polarization of the cylindrical cavity is opposite to that of the dioelectric
THe electric field inside the cylinder is thus E = \frac{P}{\epsilon_{0}} \hat{z}
the average electric field inside the cylinder is thus

E = (E_{0} + \frac{P}{\epsilon_{0}}) \hat{z}

The electric displacement is D = \epsilon_{0}E + P = \epsilon_{0}(E_{0} + \frac{P}{\epsilon_{0}}) \hat{z} -P = \epsilon_{0} E_{0}

is that correct??
thanks for the help

 

[siddharth]

so now we have two circular caps which are parallel with opposite charge densities. SO the electric field is just like that of a parallel plates each wit ha cahrge density of P

If the cavity is very small, you could treat is a parallel plate. Else, you could treat is as two discs.