# Calculate the electric potential and field

1. Sep 25, 2007

### jlucas134

Here is the question. a hollow, thin walled insulating cylinder of radius b and height h has charge Q uniformly distributed over its surface. Calculate the electric potential and field at all points along the z axis of the tube.

Outside the tube
Inside the tube.

I know how to find the field, its just -"del" V, but my problem is finding V...

I know you have to take into account the area of the surface and the radius b...

here is what I have for the integral, which i don't know is right or not. Any help would be outstanding...If someone could help me set it up, I think i could get it from there.

(Q*k )/h * int (1/R), dz, limit from 0 to h, where R is equal to sqrt(b^2+(p-z)^2)

after integration

I get a

(Q*k )/h ln [(sqrt(b^2+(p-z)^2)+h-p)/(sqrt(b^2+p^2)-p)}

2. Sep 26, 2007

### Reshma

I think you should divide your cylinder into elementary circular slabs. Find the expression for the potential at a point above the centre for one slab and integrate it for the entire length of the cylinder. Try and see if this works...

Last edited: Sep 26, 2007
3. Sep 26, 2007

### jlucas134

I tried it with no success..

I attempted to treat it like a ring of charge...finding E then integrating to find V.
Still no success.

I think I am getting lost in the "point any where on the axis inside or outside the tube".
any suggestions how to solve this?

4. Sep 27, 2007

### Reshma

The charge is distributed over the surface, so by Gauss's law the electric field inside the cylinder is zero. Find the value of the surface charge density $\sigma$ using Gauss's law for field outside the cylinder.

The electric potential of a charged ring will be given by:
$$V = \frac{1}{4\pi \epsilon_0}\int_{ring} {dq \over r} = \frac{1}{4\pi \epsilon_0}\left(\frac{1}{\sqrt{a^2 + x^2}}\int dq\right)$$
a = h in your case.
Now write dq in terms of $\sigma$ and integrate along the z-axis.