(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Griffiths Example 8.2

Determine the net force on the northern hemisphere of a uniformly charged solid sphere of radius R and charge Q. use maxwell's stress tensor to calculate the force. (Not Gauss/Coulomb's law)

2. Relevant equations

The stress tensor is defined by

[tex] T_{ij} = \epsilon_{0} (E_{i}E_{j} - \frac{1}{2} \delta_{ij} E^2) + \frac{1}{\mu_{0}} (B_{i}B_{j} - \frac{1}{2} \delta_{IJ} B^2) [/tex]

where i and j are coordinates

3. The attempt at a solution

ok For the bowl part

[tex] da = R^2 \sin \theta d\theta d\phi \hat{r} [/tex]

[tex] \vec{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{R^2} \hat{r} [/tex]

so

[tex] T_{zx} = \epsilon_{0} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin \theta \cos\theta \cos\phi [/tex]

[tex] T_{zy} = \epsilon_{0} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin \theta \cos\theta \sin\phi [/tex]

[tex] T_{zz} = \frac{\epsilon_{0}}{2} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 (\cos^2 \theta - \sin^2 \theta)[/tex]

The net froce is in the Z direction (taking the Z axis to point stragiht up, teh flat part of this bowl lying on the XY plane...

[tex] (T da)_{z} = T_{zx}da_{x}+T_{zy}da_{y}+T_{zz}da_{z} = \frac{\epsilon_{0}}{2} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin\theta \cos\theta d\theta d\phi[/tex]

I dont understand what happened in that last step...

... i know this question seems silly.... i should know all of this thoroughly in and out at this point...

anyway why does it turn into that??

what are dax and day and daz??

did they convert into spherical coords and then 'expand' ??

thanks for your help

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# Force on a uniformly charged solid sphere

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