# Electric Field of a Spherical Shell Cut in Half

1. Feb 1, 2012

### Airsteve0

1. The problem statement, all variables and given/known data
A metallic spherical shell of radius a is cut in half at its equator. The two halves are separated very slightly and are maintained at potentials $+V_{0}$ and $-V_{0}$. I am trying to find the electric field at the center of the sphere.

2. Relevant equations
The equation for the potential of the sphere was calculated using Laplace's equation in spherical polar coordinates using separation of variables, and was found to be as follows:

$V(r,\vartheta)$=$\sum$A$_{2m+1}$r$^{2m+1}$$P_{2m+1}(Cosθ$)

where "P" represents Legendre polynomials.

3. The attempt at a solution
Trivially I think it should be zero and taking the gradient of the above equation seems to support this for r=0, θ=Pi/2; however, I am unsure if this is correct reasoning. Any assistance would be greatly appreciated, thanks!

2. Feb 2, 2012

### rude man

The E field at the center is certainly not zero.

Obviously, the potential at the center is zero V. So in moving a unit test charge from the center to one of the hemisphere's "poles" it can be seen that the average E field must be V/R where R is the radius of each hemisphere. By symmetry the E field is directed from the + pole to the - pole.

(You can also show that the potential at the center is zero by computing the work done in bringing a test charge from infinity to the center along an extension of the sphere's diameter piercing its equator. At each point the force on the test charge is zero, canceled by the + and - hemispherical, symmetrically distributed charges. Work = 0, potential =0.)

Having said that, I am at a total loss as to computing the actual E field at the center. Maybe I'll look at solving the Poisson equation also. Let you know ..... sorry no one else has responded.

EDIT - see physics.usask.ca/~hirose/p812/notes/Ch2.pdf Example 3. This is obviously a toughie!

Last edited: Feb 2, 2012
3. Feb 2, 2012

### Airsteve0

thank you, I appreciate it :)

4. Feb 2, 2012

### vela

Staff Emeritus
Take the gradient and then set r to 0. Show us what you get for the gradient.

Last edited: Feb 2, 2012
5. Feb 3, 2012

### Airsteve0

I've actually already handed in my homework so I don't have my answer with me anymore. However, when I took the gradient I found that I had two terms (mutiplied by an ugly constant out front with double factorials and such) which was something like:

contant * [(2m+1)*r^(2m)*P(Cos(theta)) + r^(2m+1)*P'(Cos(theta))]

For the spherical shell cut in half, to find the electric field at the center I set r=0 and theta = Pi/2. For these conditions I said that the only m-value for which the series is not directly zero is m=0. Therefore, r=0 only kills the second term. However, theta=Pi/2 kills the first term becuase P(Cos(Pi/2))=0. There must be an error in my math or reasoning somewhere, I just dont know where it is.

6. Feb 3, 2012

### vela

Staff Emeritus
It looks like you used the incorrect expression for the gradient. In spherical coordinates, it's given by
$$\nabla V = \frac{\partial V}{\partial r}\hat{r} + \frac{1}{r}\frac{\partial V}{\partial \theta} \hat{\theta} + \frac{1}{r \sin\theta} \frac{\partial V}{\partial \phi} \hat{\phi}$$ You don't just differentiate with respect to each variable, like you do with Cartesian coordinates. Also, your expression has no unit vectors.

The first term in the expansion is $a_1 r P_1(\cos \theta) = a_1 r\cos \theta$, so you should be able to find that the first term in the gradient turns out to be $a_1(\cos\theta \,\hat{r} - \sin\theta\,\hat{\theta})$, which is equal to $a_1 \hat{z}$. The rest of the terms vanish when r=0.

Last edited: Feb 3, 2012
7. Feb 3, 2012

### Airsteve0

ugh what a silly error I made. Thank you for the clarification.

8. Feb 4, 2012

### PhysicsGente

It looks to me that the equation for the potential suggests that it doesn't vanish as r goes to infinity? How is that possible? I thought it went like 1/r ;(.

9. Feb 4, 2012

### vela

Staff Emeritus
Airsteve0 found the interior solution, which is only valid for r < R.