# Where's the logic in this Boundary value problem?

## Main Question or Discussion Point

Considering the classic problem in Electrodynamics "Conducting sphere with Hemispheres at different potentials"

How does one think in order to attack this problem? I didn't get it. What potential was considered in solving this problem? Was it the +V or the -V? Or both? Why is θ' considered from 0 to ∏/2?

Here's the link: www.physics.hku.hk/~phys6503/Chapter3ZDW.ppt [Broken] (Slide number 16)

Thank you.

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Simon Bridge
Homework Helper
Why make everyone download an entire ppt presentation just for a few of the slides? We are not all on cheap ultra-fast broadband :(

Please separate out what you want us to look at and put in an accessible format... you could have linked in the pic and wrote out the equations for example, or exported the slides to pdf format.

Anyway - having already been burned:
I see the problem you wanted starts at slide 17, like this:

$$V(r,\theta,\varphi)=\frac{V}{4\pi}\int_0^{2\pi} d\phi^\prime \bigg[ \int_0^{\pi/2} d(-\cos\theta^\prime)-\int_{\pi/2}^\pi d(-\cos\theta^\prime) \bigg] \frac{a(r^2-a^2)}{(a^2+r^2-2ar\cos\gamma)^{3/2}}$$ note: the slide uses {} instead of [] but it does mix ##\varphi## and ##\phi## and group the integrals like that. The slide does not say what "a" is either.
What potential was considered in solving this problem? Was it the +V or the -V? Or both? Why is θ' considered from 0 to ∏/2?
... Both potentials were considered - using the superposition principle. ##\theta## is the angle to the z-axis and each hemisphere occupies 90 degrees of ##\theta##... hence the ##\pi/2##.

Put away the slide and work it out by your own favorite method, then go back and compare.

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Oh, I apologize, excuse me. Thank you for your reply. I will do.

clem
"Conducting sphere with Hemispheres at different potentials" cannot happen.
It must be two separate hemispheres with perfect insulation between them.

WannabeNewton