# Potential of a hemisphere

Discussion in 'Physics' started by AlexK, May 30, 2007.

1. ### AlexK Thread Starter Active Member

May 23, 2007
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0
If i have an uniformally charged hemisphere with its center at the origin, how do i calculate the Electrostatic potential (or the field) at each point?

In the case of a whole sphere it's easy to use Gauss' law
due to the spherical symmetry.

But what to do in the case of a hemisphere?

thanks.
Alex.

2. ### Papabravo Expert

Feb 24, 2006
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Is the hemisphere a closed surface or not? Is there a disc which goes through the origin and intersects the hemisphere? If not then this problem is similar to the charge on one side of a flat plate. Where are the charges by the way?

3. ### AlexK Thread Starter Active Member

May 23, 2007
34
0
The hemisphere is an open surface.
How is it similar to the charge on one side of a flat plate?
The charge is uniformally distributed on the hemisphere.

4. ### Papabravo Expert

Feb 24, 2006
11,024
2,119
In the flat plate you use an image charge to visualize the field lines. Then the whole surface is at the same potential, do I have this correct?

5. ### AlexK Thread Starter Active Member

May 23, 2007
34
0
I'm not really following,
don't you need a conductor in the vicinity of some charge distribution in order to use the method of images? And how can you tell right away that this is similar to the charge on one side of a flat plate?

I suppose we need to divide the hemisphere to incremental charges dq, and using super-position find the total potential, but i get mixed up in the integrals and the geometry of the problem.

thanks.

6. ### Papabravo Expert

Feb 24, 2006
11,024
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Oh now I see, the charges are distributed on the hemisphere. So is the hemisphere a conductor? Are the charges free to move about on the hemisphere? Is there any reason that they should - given the geometry of the problem?

7. ### recca02 Senior Member

Apr 2, 2007
1,211
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sorry for interrupting,
potential at what point?
are u looking for a generalized expression?
i was actually thinking if it is to be calculated at origin;
summing up the scalar quantity potential would it not equal = Q/R (since each charge is at distance- Radius and total charge adss up to Q)
(sorry if i m getting it wrong , i only touched this subject once)

8. ### Papabravo Expert

Feb 24, 2006
11,024
2,119
At large distances the geometry near the origin does not matter. The Electric Field will be along rays from the origin to infinity. The equipotential surfaces will be concentric shells.

If we assume that the great circle of the hemisphere sits on the xy plane and it is the "Northern" hemisphere with the nort pole in the positive z-direction the it takes more work to approch from infinity in the minus z-direction then it does to approach the north pole from infinity in the positive z-direction.

That's as much intuition as I can muster at the moment, but I'll keep thinking about it.

9. ### AlexK Thread Starter Active Member

May 23, 2007
34
0
The hemisphere is a conductor, so i guess they are free to move but the charge distirbution is already given so i dont think it should matter.

I was looking for the generalized expression but i guess it would be hard to find it anywhere except the symmetry axis.
I think you are right about the potential at the origin it does equal Q/R.

And i think i know how to solve for all the other points on the z axis, if we divide the hemisphere into charges hoop of thickness Rd(teta) each with radius Rsin(teta)
find the contribution of each hoop to the potential on some point on the axis and then integrate over the entire sphere.
teta - the angle from the z axis.

10. ### Irid New Member

Jan 27, 2007
4
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The idea is first to solve a simpler problem. Say, upper hemisphere has charge $+Q/2$ and the lower has $-Q/2$. After solving for this potential you simply superimpose another sphere having a uniform charge $+Q/2$.

Now how do you solve the simpler problem, in the first case? You can do this by solving Laplace equation in spherical coordinates, so that
$\begin{eqnarray*}
\Phi_{\textrm{in}} &=& \sum_{l=0}^{\infty} A_l r^l P_l(\cos \theta)\\
\Phi_{\textrm{out}} &=& \sum_{l=0}^{\infty} B_l r^{-(l+1)} P_l(\cos \theta)\\
\end{eqnarray*}$

The two solutions must agree on the surface $r=a$, while the expansion coefficients can be found via
$\sigma(\theta) = \epsilon_0 \frac{\partial \Phi}{\partial r} |_{r=a}$
Using these guidelines I found the potential to be
$\Phi_{\textrm{in}} = \frac{Q}{2\pi \epsilon_0 a} \sum_{l\textrm{ odd}}^{\infty} \left(\frac{r}{a}\right)^l \left(-\frac{1}{2}\right)^{\frac{l-1}{2}} \frac{(l-2)!!}{2\left(\frac{l+1}{2}\right)!} \frac{2l+1}{2l} P_l(\cos \theta)$
and
$\Phi_{\textrm{out}} = \frac{Q}{8\pi \epsilon_0 r} + \frac{Q}{2\pi \epsilon_0 a} \sum_{l\textrm{ odd}}^{\infty} \left(\frac{a}{r}\right)^{l+1} \left(-\frac{1}{2}\right)^{\frac{l-1}{2}} \frac{(l-2)!!}{2\left(\frac{l+1}{2}\right)!} \frac{2l+1}{2l} P_l(\cos \theta)$

Similiar problem is considered in Jackson's Classical Electrodynamics, Section 3.2.