- #1

Jazz

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## Homework Statement

Calculate the Electric Field generated by a hemisphere of radius ##R## (and uniform charge density ##\sigma##) at its center.

## Homework Equations

##\displaystyle E = \frac{Q}{4\pi \varepsilon_0 r^2}##

## The Attempt at a Solution

I think I'm doing some mistakes with the math. When trying to find the field by spherical coordinates I get something different if I try by using rectangular coordinates.

Attempt:

##\displaystyle \boxed{\mathrm{d}E = \frac{\mathrm{d} Q}{4\pi \varepsilon_0 R^2}}##

- Spherical Coordinates:

##\displaystyle \mathrm{d}Q = \sigma \mathrm{d}S##

##\displaystyle \mathrm{d}Q =\sigma R^2 \sin\phi \mathrm{d}\phi \mathrm{d}\theta##

By symmetry ##\mathrm{d}E_y = \mathrm{d}E_z = 0## and ##\mathrm{d}E = \mathrm{d}E_x##, so

##\displaystyle E = \frac{\sigma}{4\pi \varepsilon_0}\int\limits_{\pi / 2}^{3\pi / 2}\int\limits_{0}^{\pi}\sin\phi \mathrm{d}\phi \mathrm{d}\theta = \frac{\sigma}{2\varepsilon_0}##

- Rectangular Coordinates:

##\displaystyle \mathrm{d}Q= 2\sigma\pi R^2\sin\alpha \mathrm{d}\alpha##

Again we apply symmetry. Hence,

##\displaystyle \mathrm{d}E_x = \mathrm{d}E \cos\alpha##

##\displaystyle E = E_x = \frac{\sigma}{2 \varepsilon_0}\int\limits_{0}^{\pi /2}\sin\alpha\cos\alpha \mathrm{d}\alpha = \frac{\sigma}{4\varepsilon_0}

##

There is a difference of a factor of 2 between the first and the second way. I don't know what I'm missing :/.

Thanks!

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