Obtaining electric field of an uniformly charged sphere surface without using gauss law [closed]

Question

How can i obtain the electric field due to a uniformly charged sphere surface without using gauss law on a point outside the sphere, im stuck not knowing what infinitesimal surface i shall consider so that i can integrate all over the sphere surface.

Answer 1

The charge distribution when the charge is only on the surface is given by: $$ \varrho(r,\varphi,\theta) = \frac{Q}{4\cdot\pi\cdot r^2} \cdot \delta(R - r)$$ where $Q$ is the total amount of charge on the sphere with radius $R$ and $\delta(R - r)$ is the Dirac-Delta. Now use $$ \varphi(\vec{r}) = \frac{1}{4\pi\epsilon_0} \int \frac{\varrho(\vec{r}\ ')}{|\vec{r} - \vec{r}\ '|} d^3r' $$ while one can here assume that $\vec{r} = r \cdot \vec{e}_z$ because the problem is spherical symmetric. Therefore: $$ \frac{1}{|\vec{r} - \vec{r}\ '|} = \frac{1}{\sqrt{r^2 + r\ ' \ ^2 - 2rr\ ' \cos \theta \ '}} $$ This leaves the integration quite easy: \begin{align} \varphi(\vec{r}) & = \frac{1}{4\pi\epsilon_0} \int_{0}^{\pi} \int_{-\infty}^{\infty} \int_{0}^{2\pi} \frac{\varrho(\vec{r}\ ')}{|\vec{r} - \vec{r}\ '|} r\ ' \ ^2 \sin\theta \ ' ~ d\varphi\ ' ~ dr\ ' ~ d\theta\ ' \\ & = \frac{Q}{8\pi\epsilon_0} \int_{0}^{\pi} \int_{-\infty}^{\infty} \frac{\delta(R - r\ ')}{\sqrt{r^2 + r\ ' \ ^2 - 2rr\ ' \cos \theta \ '}} \sin\theta \ ' ~ dr\ ' ~ d\theta\ ' \\ & = \frac{Q}{8\pi\epsilon_0} \int_{-1}^{1} \frac{1}{\sqrt{r^2 + R^2 - 2rR \cos \theta \ '}} ~ d(\cos\theta\ ') \\ & = \frac{Q}{8\pi\epsilon_0} \cdot \left[- \frac{\sqrt{r^2 + R^2 - 2rR x}}{rR} \right]^{x=1}_{x=-1} \\ & = \frac{Q}{8\pi\epsilon_0rR} \cdot \left( \sqrt{(r+R)^2} - \sqrt{(r-R)^2} \right) \\ & = \frac{Q}{4\pi\epsilon_0} \cdot \left[ \matrix{\frac{1}{r} \ \ \textrm{ for } \ r > R \\ \frac{1}{R} \ \ \textrm{ for } \ r < R} \right] \end{align} Now take the gradient to get the electro-static-field:

$$ \vec{E} = - \nabla \varphi = \frac{Q}{4\pi\epsilon_0} \cdot \left[ \matrix{\frac{1}{r^2} \cdot \frac{\vec{r}}{r} \ \ \textrm{ for } \ r > R \\ 0 \ \ \textrm{ for } \ r < R} \right] $$

Answer 2

The area element for a sphere of radius $R$ in spherical coordinates is $$dA = R^2 \sin\theta\ d\theta \ d\phi$$ and $dq = \sigma\ dA$. where $\sigma$ is the surface charge density.

Your differential E-field element is $$d\vec{E} = \frac{k\ dq\ \hat{r}}{r^2}$$ where $\hat{r}=\vec{r}/r$ is a function of the angles.

You will need to determine the vector position $\vec{r}$ from each surface element (as a function of $\theta$ and $\phi$) to the point of interest in space (where you want to know the electric field). You do not integrate over $r$.