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Suppose we have a spherical surface with a surface charge density varying as $cos(\theta)$. Apparently one can find the electric field both outside and inside such a spherical surface by superposing the fields of two slightly offset charged spheres with uniform volume charge density. Tips regarding how one goes about doing such a thing would be greatly appreciated. |
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To be honest, I just learned about all this myself in the last months, so I am not sure whether this is actually correct. Since you have this spherical symmetry, I think that you need spherical harmonics. They are orthogonal functions, think of them as a Fourier series on the surface of a sphere. Your charge density $\sigma$ does not depend on $\phi$, therefore we can use the simpler Legendre-polynomials, where $m = 0$. Fist, we want to express the potential like so: $$ \varphi(\vec x) = \frac1{\varepsilon_0} \sum_{l=0}^\infty \frac{1}{2l+1} \varrho_{l} \frac{1}{r^{l+1}} Y_{l,0} (\theta, \phi) $$ The coefficients $\varrho_l$ are given by: $$ \varrho_l = \int \mathrm d^3 x'\, r'^l \varrho(\vec x) Y^*_{l,0} (\theta', \phi') $$ Now with $$ Y^*_{l,0}(\theta', \phi') = \sqrt{\frac{2l+1}{4\pi}} P_l(\cos \theta') $$ and $$ P_0(x) = 1 ,\quad P_1(x) = x $$ we can calculate the coefficitents. But first, we need to convert the surface charge density $\sigma$ into a volume charge density. For that, we use the $\delta$-distribution: $$ \varrho(\vec x) = \delta(r-R) \sigma_0 \cos(\theta) $$ If you plug those into the $\varrho_l$, you will get $\varrho_0 = 0$ and $\varrho_1 = \sqrt{\frac{3}{4 \pi}} \frac{4}{3} \pi R^3 \sigma_0$. I hope this is correct. Then we can put this into the first formula and get $\varphi$: $$ \varphi(\vec x) = \frac{1}{\varepsilon_0} \sqrt{\frac{3}{4 \pi}} \frac{4}{3} \pi R^3 \frac{1}{r^2} \sigma_0 \cos \theta $$ Since this is a pure dipol potential, the $1/r^2$ seems about right. And if you look at the dimensions, the $R^3 \sigma_0 / r^2$ have just the needed Charge/Length. Spherical harmonics might be overkill, maybe there is a simpler method to do this. |
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