Polarization Charge on the surface of a spherical cavity

[XCBRA]

Homework Statement

The polarizatiob charge on the surface of a spherical capacitor is $$-\sigma_e \cos(\theta),$$ at a point whose radius vector from the centre makes an angle $\theta$ witha given axis Oz. Prove that the field strength at the centre is $$\frac{\sigma_e}{3 \epsilon_0},$$

Homework Equations

The Attempt at a Solution

Well I not entirly sure how to approach this problem. I tried exapanding the potentials inside and outside the sphere as:
$$V_{in} = A_1 r \cos(\theta) + \frac{A_2}{r^2}\cos(\theta) V_{out} = B_1 r \cos(\theta) + \frac{B_2}{r^2}\cos(\theta)$$

Then since $V_{in} \neq \infty,$ $A_2 = 0 ,$

then saying that at r=R: $D^{perpendicular}_{in} - D^{perpendicular}_{out} = \sigma_f$ and assuming that both inside and outside have the same permitivitty then $$E^{radial}_{in} - E^{radial}_{out} = \frac{\sigma_f}{\epsilon_0}$$.$$A_1 + B_1 - \frac{B_1}{R^2} = \frac{\sigma_e}{\epsilon_0}$$

I am not entirely sure if I am aproaching this problem in the right way. Any help or advice on how to g o about solving this problem or problems like this would very much apreciated.

 

[mathfeel]

Homework Statement

The polarizatiob charge on the surface of a spherical capacitor is $$-\sigma_e \cos(\theta),$$ at a point whose radius vector from the centre makes an angle $\theta$ witha given axis Oz. Prove that the field strength at the centre is $$\frac{\sigma_e}{3 \epsilon_0},$$

While I am sure there is some clever way to apply spherical harmonic expansion here, the problem is more straight forwardly done by just use integration. Just do a surface integral over the electric field contribution of each area element. There is clear azimuthal rotation symmetry so that you only needs to compute the $z$ component. $$E_z = \frac{1}{4\pi \epsilon_0} \int \frac{\sigma(\theta) \cos\theta \, da}{R^2} \,,$$ where $R$ is radius of the sphere.

 

[XCBRA]

Fistly thank yo for your help. Secondly I am having trouble seeing how yuo get to this integral and then how you solve it. Am I correct in saying this is a result of Gauss' Law?

I have tried to compute the integral and i am unfortunately at a loss as to how to do this to get the desired answer:

$$E_z = \frac{1}{4\pi \epsilon_0} \int \frac{\sigma(\theta) \cos\theta \, da}{R^2}$$

$$= \frac{1}{4\pi \epsilon_0} \int\int \frac{\sigma(\theta) \cos\theta \, rdrd\theta}{R^2}$$

$$= \frac{1}{4\pi \epsilon_0} \int r dr \int \frac{\sigma(\theta) \cos\theta \, d\theta}{R^2}$$

here is where i am stuck. Do I take limits, and if so how do I determine them, I have tried taking r=R to r=0 and $\theta = 0, 2\pi.$? Thank you for you help.

 

[mathfeel]

You have the wrong area element of a sphere. It should be: $da = R^2 \sin \theta d\theta d\varphi$.

There is no variation in radial distance, so no $dr$.

 

[XCBRA]

Ah ok that makes a lot more sense, thank you for your help.