Find the surface charge that appears on a cavity

Question

The problem is of a conducting medium in a uniform field $E_0$. A spherical cavity of radius $a$ is formed inside said medium.

I already found the potential inside and outside the cavity

$\Phi_{\text{int}}=- \frac{3}{2}E_0r\cos(\theta)$

$\Phi_{\text{ext}}= -E_0r\cos(\theta)-\frac{1}{2}E_0a^3\frac{\cos(\theta)}{r^2}$

But I can't find a way to find the surface charge density on the cavity.

It occurs to me to take the value of the potential $\Phi_{\text{int}}$ and calculate the electric field using $E=-\nabla \Phi_{\text{int}}$ and then use Gauss's law to find $\sigma$ but it doesn't work out the result. The book tells me that the result for this density should be

$\sigma=-\frac{3}{2}KE_0\cos(\theta)$

But I don't know who that "$K$" is. I guess it's the dielectric constant which is defined as $K=\epsilon/\epsilon_0$. If it is the dielectric constant, I don't know where to get it from. Can you give me any suggestion?

Answer 1

I already found the potential inside and outside the cavity

$\Phi_{\text{int}}=- \frac{3}{2}E_0r\cos(\theta)$

$\Phi_{\text{ext}}= -E_0r\cos(\theta)-\frac{1}{2}E_0a^3\frac{\cos(\theta)}{r^2}$

Integrating $\vec \nabla \cdot \vec E = \rho/\epsilon_0$ over a pillbox (small cylinder) oriented with its long direction along $\hat r$, we find: $$ \int_{V_{pillbox}}dV \vec \nabla \cdot \vec E = Q_{enc}/\epsilon_0 = \int_{boundary}\vec {dS}\cdot \vec E = \delta S(E_{ext}^r(r=a) - E_{int}^r(r=a)) $$ So, that: $$ \sigma/\epsilon_0 = -E_{int}^r(r=a)\;, $$ since $E_{ext}^r(r=a)=0$.

Or: $$ \sigma = \epsilon_0 \frac{\partial \Phi_{int}}{\partial r} = - \frac{3}{2}\epsilon_0E_0\cos(\theta) $$

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Another way to approach this problem is to open up your copy of Jackson's "Classical Electrodynamics" (3rd Edition) to page 18 and see that: $$ (\vec D_2 - \vec D_1)\cdot \hat n = \sigma\;, $$ but in our case, we have $\vec D_2\cdot\hat n = 0$ at the boundary (per your equations). And we have $-\vec D_1\cdot\hat n = -\epsilon_0E_{int}^r$, since the cavity is empty.

So, again we find: $$ \sigma = -\epsilon_0E_{int}^r $$