# Finding electric field at the center of a polarized cylinder

1. Nov 6, 2015

### phys-student

1. The problem statement, all variables and given/known data
Consider a uniformly polarized electret in the shape of a cylinder with height h and radius 10h. The polarization in the dielectric is P, parallel to the cylinder axis.
a) calculate surface and volume bound charges
b) calculate the electric field at the center of the cylinder. Since the radius is large compared to the height, you may neglect edge effects.

2. Relevant equations
σb = P⋅n
ρb = - ∇⋅P

3. The attempt at a solution
I've already worked out part a and found there's a surface bound charge of P on the upper surface, and -P on the lower surface. I tried to do part b by finding the potential in the middle of the cylinder due to the top and bottom surface separately, then adding them together to get the total potential, then taking the gradient to find the electric field. When I tried to do this the potential from the top and bottom surfaces cancel giving me 0 total potential and thus no electric field, but I know this doesn't make any sense since there should be field lines going from the positive surface to the negative surface. Can someone help point me in the right direction?

2. Nov 6, 2015

### gleem

Zero potential does not imply zero field, is that not correct? Why?

3. Nov 6, 2015

### phys-student

I was trying to use the equation:
E = -∇V
To find the electric field. If the potential is 0 then the gradient will also be 0, this is why I thought that no potential means there is no electric field

4. Nov 6, 2015

### gleem

What is the interpretation of the term gradient?

5. Nov 6, 2015

### phys-student

Just the normal definition of the gradient. Derivative w.r.t. x times x-hat plus derivative w.r.t. y times y-hat plus derivative w.r.t. z times z-hat

6. Nov 6, 2015

### gleem

So what does the value of a function at a point have to do with its gradient?

7. Nov 6, 2015

### phys-student

This is the approach that I used:

I wanted to find an expression for the potential due to the top and bottom discs separately, then sum the 2 expressions to find the total potential, once I have an equation for the total potential I would take the gradient of it to find the electric field. However, when I try to sum the 2 expressions I am left with 0 since the geometry of the discs is the same but the charges are opposite. If there is something wrong with the approach I'm using please let me know

8. Nov 6, 2015

### gleem

What is the equation for the potential that you are using?

9. Nov 6, 2015

### phys-student

I used the equation for finding potential due to a surface charge:
V = (1/4*pi*ε) ∫ (σ/r2) da
Where ε is the vacuum permittivity constant, σ is the surface charge density, and r is the distance between the disc and the field point.
Evaluating this expression for the top disc I get:
Vtop = (σ/2*ε)*[(r2 + z2)0.5 - z]
Where r is the radius of the disc. When I do the same thing for the bottom disc I get the same expression except with -σ instead of σ. So when I try to sum them I get 0.

10. Nov 6, 2015

### gleem

Ah ha. You can't use the same general expression for the potential for both sides of the disk as in general you will be closer to one surface compared to the other..

dVtop ~ 1/ (r2 + z2)½ while dVbot ~ 1/ (r2+(t-z)2)½
where r is the distance from the center to the position of an elemental area da, z is the distance from the top and t is the thickness of the disk.

??

11. Nov 6, 2015

### phys-student

Sorry i must not have made the problem clear enough... I am looking for the potential in the center of a uniformly polarized cylinder. Since the polarization causes surface charges to only exist on the top and bottom surfaces, I was approximating the situation by looking for the potential between 2 discs with surface charge densities that are equal in magnitude but with opposite signs. The expression you've given me appears to be for the potential on either side of a disk with a finite thickness which is not what I'm looking for

12. Nov 6, 2015

### gleem

You solve the problem for the general case i.e., the potential anywhere on the central axis, take the gradient of that equation then evaluate the result at the center of the cylinder. the electric field depend on how the potential varies on the central axis. You want to express the distance from the front and the back with the same variable ( say z). From the front it is z but from the back it is the thickness -z. the acutal potential at the midpoint will be zero by the function will not vanish so you can take the gradient of it. and get E.

13. Nov 6, 2015

### phys-student

Okay now I get what you're saying. Thanks for the help, just finished the problem now