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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!

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    $\begingroup$ The issue is the integral is improper. To evaluate it, remove an $\epsilon$-ball from $S$ and compute the resulting integral. Then take the limit as $\epsilon \to 0$. $\endgroup$ – kobe Feb 14 '15 at 22:10
  • $\begingroup$ @kobe still there is a term $\ln R$, which shouldn't be part of the answer: $(\sigma R /4\epsilon_0,0,$anything$)$ $\endgroup$ – Vladimir Vargas Feb 14 '15 at 22:17
  • $\begingroup$ According to my calculations, $\ln R$ is not present. It is indeed $\frac{\sigma R}{4\epsilon_0}\vec{k}$. $\endgroup$ – kobe Feb 14 '15 at 22:21
  • $\begingroup$ If I understood well, generating this $\epsilon$-ball is changing the limits for $\rho'$: $\epsilon\le\rho'\le R$ right? $\endgroup$ – Vladimir Vargas Feb 14 '15 at 22:23
  • $\begingroup$ Yes. I'll post my calculations up soon. $\endgroup$ – kobe Feb 14 '15 at 22:24
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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}

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  • $\begingroup$ Yes, I did not consider r hat that way $\endgroup$ – Vladimir Vargas Feb 14 '15 at 22:57
  • $\begingroup$ kobe, wouldn't the unit vector from a point $P$ in $S$ to the origin be $-\langle\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\theta\rangle$? where $\phi$ is the angle between the positive $z$ axis to the segment $\overline{OP}$ from the origin to the point and $\theta$ the angle between the positive $x$ axis and the projection of $\overline{OP}$ onto the $xy$ plane $\endgroup$ – Vladimir Vargas Feb 15 '15 at 3:29
  • $\begingroup$ The minus sign should not be present: $\hat{\mathbf{r}}(x,y,z)$ is the vector from the origin to $(x,y,z)$, i.e., $\langle x,y,z\rangle$. In spherical coordinates, this is $\langle \sin \phi\cos\theta, \sin\phi\sin\theta,\cos\phi\rangle$. $\endgroup$ – kobe Feb 15 '15 at 3:35
  • $\begingroup$ @Vladmir please let me know more specifically what issue you have with my answer so I can address it. $\endgroup$ – kobe Feb 15 '15 at 3:41
  • $\begingroup$ @Vladmir putting the minus sign aside, the $z$-component you have should be $\cos \phi$, not $\cos \theta$. Think of it this way. If you were to take the norm of $-\langle \sin \phi\cos \theta, \sin \phi\sin \theta, \cos \theta\rangle$, you would get $\sqrt{\sin^2\phi + \cos^2\theta}$, which does not equal $1$. $\endgroup$ – kobe Feb 15 '15 at 3:49
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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
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  • $\begingroup$ here is a mathjax guide to type maths on this site. $\endgroup$ – Siong Thye Goh Jan 30 '19 at 17:36
  • $\begingroup$ We don't all know the result for a disk. $\endgroup$ – sammy gerbil Feb 23 '19 at 12:49
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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.

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  • $\begingroup$ I took $\hat{\mathbf r} = - \dfrac{(\rho',\phi',\theta')}{\rho'}$. $\endgroup$ – Vladimir Vargas Feb 14 '15 at 22:10
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    $\begingroup$ Indeed, but this is still not right. Consider the vector $\hat{{\bf r}}$ in cartesian coordinates, this is $\hat{{\bf r}} = \frac{x \hat{\bf i} + y \hat{\bf j} +z \hat{\bf k}}{\sqrt{x^2 + y^2 + z^2}}$. In spherical coordinates the vector $\hat{\bf r}=\hat{\bf e}_{r}$ does not have the same expression, it is a basis vector. What you probably need for your calculation is to resolve $\hat{\bf e}_r$ in cartesian coordinates and see that only the $z$-component survives $\endgroup$ – Rogelio Molina Feb 14 '15 at 23:13

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