Volume integral of electric field (hemisphere solid)

Question

Let $S$ be a hemisphere of radius $R$, and let $\sigma$ be the constant charge density at each point $(x',y',z')$ in $S$. The electric field generated by the hemisphere is a vector function: $$\mathbf E(x,y,z)=\frac{1}{4\pi\epsilon_0}\int_S\frac{\sigma}{r^2}\hat{\mathbf r}\,dV,$$

Where $\hat{\mathbf r}$ is the unit vector from a point $(x',y',z')\in S$ to $(x,y,z)$, and $r^2$ is the squared distance from $(x',y',z')\in S$ to $(x,y,z)$. Consider the transformation from spherical coordinates to rectangular coordinates. Then, if we want to calculate the electric field at the centre $O$ of the hemisphere, i.e. the centre of the biggest circle that it contains, we would want to let the origin of the coordinate system to be at $O$. Once we do that, the electric field calculated at $O$ is:

$$\mathbf E = \frac{\sigma}{4\pi\epsilon_0}\int_0^{2\pi}\int_0^{\pi/2}\int_0^R-\frac{(\rho',\phi',\theta')}{\rho'}\sin\phi\,d\rho'd\phi'd\theta'.$$

This integral will produce a $\ln\rho'|_0^R$, how do I deal with $\ln 0$? Did I commit any mistake? I appreciate your help in pointing out what could be wrong.

PS: I know I can solve the problem partitioning the hemisphere into rings, but I want this solution, thanks!

Answer 1

This integral is improper, so to evaluate it, we remove a $\delta$-ball inside $S$, convert to spherical coordinates, then take the limit as $\delta\to 0$. The result is

\begin{align} \mathbf{E} &= \lim_{\delta \to 0} \frac{1}{4\pi \epsilon_0}\int_0^{2\pi}\int_0^{\pi/2}\int_\delta^R \frac{\sigma}{r^2} \hat{r} r^2\sin \phi \, dr\, d\phi\, d\theta\\ &= \frac{\sigma}{4\pi\epsilon_0} \lim_{\delta\to 0} \int_0^{2\pi} \int_0^{\pi/2} \int_\delta^R \hat{r}\sin \phi\, dr\, d\phi\, d\theta\\ &= \frac{\sigma}{4\pi \epsilon_0} \lim_{\delta \to 0} \int_0^{2\pi} \int_0^{\pi/2} \int_\delta^R \langle\cos \theta\sin\phi, \sin\theta\sin \phi, \sin \phi\rangle\sin \phi\, dr\, d\phi\, d\theta\\ &= \frac{\sigma}{4\pi \epsilon_0} \lim_{\delta\to 0} \int_0^{2\pi} \int_0^{\pi/2} (R - \delta)\langle\cos\theta\sin^2\phi, \sin\theta\sin^2\phi, \sin^2\phi\rangle\, d\phi\, d\theta\\ &= \frac{\sigma R}{4\pi \epsilon_0} \lim_{\delta\to 0} (R - \delta) \int_0^{2\pi} \int_0^{\pi/2} \langle \cos \theta\sin^2\phi, \sin\theta\sin^2\phi, \sin^2\phi\rangle\, d\phi\, d\theta \\ &= \frac{\sigma R}{4\pi \epsilon_0} \left\langle 0,0,2\pi \cdot \frac{1}{2}\right\rangle\quad (\text{since $\int_0^{2\pi} \sin\theta \, d\theta = \int_0^{2\pi}\sin \theta\, d\theta = 0$, $\int_0^{2\pi}\sin^2\phi\, d\phi = 1/2$})\\ &= \frac{\sigma R}{4\epsilon_0}\vec{\mathbf{k}}. \end{align}

Answer 2

I think it can be done in a simple way,

    we all know that electric field created by disc of surface charge density 

      E=(rho)/2ϵ0 (1-cosθ)   

so dividing the hemisphere into smaller discs of surface charge density (σ)(dz) and integrating the integral from zero to R.

 dE=  ((σ)(dz)/2ϵ0) (1-cosθ) 
 dE=  (σ/2ϵ0) (dz-zdz/R)      since Rcosθ=z

          ∫dE= (σ/2ϵ0) ( ∫dz-∫(zdz/R))   limitsfrom 0 to R

          E=  (σ/2ϵ0) ( R-R/2)

         E=  σR/4ϵ0

Answer 3

The problem is that the ${\bf r}$ vector is not $(\rho', \phi', \theta')$, rather it is ${\hat{\bf e}}_r$, hence there is no logarithm.