(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The polarization charge on the surface of a spherical cavity is [tex] - \sigma_e cos(\theta) [/tex] at a point whose radius vector from the centre makes an angle [tex] \theta [/tex]with a given axis Oz. Prove that the field strength at the centre is [tex] \frac {\sigma_e}{3 \epsilon_o} [/tex] parallel to Oz.

If the cavity is in a uniform dielectric subject to a field of strength [tex] E_o [/tex] parallel to the direction [tex] \theta = 0 [/tex], show that [tex] \sigma_e = \frac {3 E_o \epsilon_o ( \epsilon_r -1)} {1 + 2 \epsilon_r} [/tex] where [tex] \epsilon_r [/tex] is the relative permittivity of the dielectic. Verify that that this gives the correct value for the field strength at the centre of the cavity [tex] E = \frac {3 \epsilon_r E_o} {1 + 2 \epsilon_r}[/tex] and note that [tex] \sigma_e [/tex] is not simply [tex] ( \epsilon_r - 1 ) \epsilon_o E_o [/tex] because of the distortion of the field in the dielectric caused by the presence of the cavity.

3. The attempt at a solution

The first part of the question is no problem and deriving the field strength in the center which it asks you to verify using boundary conditions etc. etc. is no problem.

However polarization charge density = [tex] P . \hat{n} = - P cos( \theta ) [/tex] which is the form of what's given at the beginning of the question leading to [tex] \sigma_e = P [/tex]

Therefore:

[tex] P = ( \epsilon_r - 1 ) \epsilon_o E [/tex]

Subbing in my equation for field inside:

[tex] P = ( \epsilon_r - 1 ) \epsilon_o \frac {3 \epsilon_r E_o} {1 + 2 \epsilon_r}[/tex]

Which is the answer they want, apart from I have an extra [tex] \epsilon_r [/tex] term in the numerator.

Any ideas?

Many thanks in advance.

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# Homework Help: Spherical Cavity in Dielectric Subject to Uniform Field

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