Potential energy of an hemispheric shell

1. Oct 9, 2011

BSJ90

1. The problem statement, all variables and given/known data
Given that the hemisphere has a charge +Q distributed through its surface with radius a. Find the electric potential on any point on z axis (the plane of the hemisphere is oriented in positive z direction).

2. Relevant equations
$\phi$ = $\int$$\frac{kQ}{|R-R'|}$*ds
(surface integral)

3. The attempt at a solution
So i decided to break this problem into little pieces of the surface using some trig.

I made a angle θ from the plane of the hemisphere around the cross section (so there is semi circles) and then an angle $\varphi$ on the xy-plane.

I found ds to be a$^{2}$d$\varphi$d$\theta$ (using the arc lengths of my angles and radii).

Now my problem is finding an equation for the vector R-R'. I tried making it a function of $\theta$ (i.e. i got (sqrt(z$^{2}$ + a$^{2}$sin$^{2}$($\theta$)
but the integral was not a nice one and it lead me to believe that it wasn't correct. if someone can help me along that would be great!

2. Oct 10, 2011

ehild

Is the charge evenly distributed on the surface?

Make a picture and indicate R and R'!

ehild

3. Oct 10, 2011

BSJ90

oh yeah evenly distributed on surface. I made a paint object to help clarify. R' is to the ds and R is to the point on the z axis. I know R has to be z but im having a problem figuring out R'. Thanks for this and the last problem you help me with.

Attached Files:

• Untitled.png
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4. Oct 10, 2011

ehild

Nice picture! :rofl: You need the magnitude of the vector difference of R-R'. What about using the Law of Cosines?
To make your task more clear I draw an other picture with Paint (using the option to draw circle, ellipse, straight line )

ehild

Attached Files:

• hemisphere.JPG
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5. Oct 10, 2011

BSJ90

ok so should my theta be from the z axis and i think that changes my whole set up oh wait i can use 90 + theta and it would work. Ok thanks time to go to work on it. Plus i only really care about the z direction.

6. Oct 10, 2011

BSJ90

I get my |R- R'| = $\sqrt{ a^2 +z^2 + 2azsin(θ)}$. look right? Umm this isn't turning out to be a easy integral. Im thinking mabye go back to Cartesian.

Last edited: Oct 10, 2011
7. Oct 10, 2011

ehild

Last edited: Oct 10, 2011
8. Oct 10, 2011

BSJ90

I just need to add 90 degrees to my theta so it restricts it to my hemisphere not a whole sphere. The ds stays the same except for that sin(θ) term which i added.

9. Oct 10, 2011

ehild

It will not be sin(theta) in the surface element of your special spherical system of coordinates.

ehild