# 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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Views:
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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