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## Homework Statement

Using direct integration, compute the electric field at the point (0,0,Z) of a semi-espherical (z > 0) charge distribution of radius R < Z and density ##\rho=constant## having a spherical hole of radius ##r_h<R## centered at (0,0,##r_h##).

## Homework Equations

Coulomb's law in differential form:

## d\vec{E} = \frac{Kdq}{|\vec{r} - \vec{r}_{dq}|^2}\frac{\vec{r} - \vec{r}_{dq}}{|\vec{r} - \vec{r}_{dq}|}##

##\vec{r} = Z \hat{k} ## (the field point)

##\vec{r}_{dq} = r\sin(\theta)\cos(\phi)\hat{\imath} + r\sin(\theta)\sin(\phi)\hat{\jmath} + r\cos(\theta)\hat{k}## (the source point)

## r = 2r_h\cos(\theta) ## (the spherical hole in spherical coordinates )

## The Attempt at a Solution

By symmetry, the only relevant integral is:

## \vec{E}(z) = 2\pi K \rho \int_{\theta=0}^{\theta=\pi/2} d\theta \int_{r=2 r_h \cos{\theta}}^{r=R} dr \left[ \frac{ Z-r\cos(\theta) }{ (r^2 - 2 r Z \cos(\theta) + Z^2)^{3/2} } \right]r^2 \sin(\theta) ##

**The question is how to perform the integral**. Using brute force via the wolfram alpha site

finished with

*Standard computation time exceeded...*

https://www.wolframalpha.com/input/...,+{r,+2*rh*Cos[theta],+R}],+{theta,+0,+Pi/2}]

Thanks in advance for your help,

Sergio

PS. By the way, the formulation seems to be right as one can get the right ##\vec{E}## field of the full semi-sphere