# Potential Problem

1. Mar 15, 2013

### Miike012

I would like to know if there is a way of finding the Potential of a disk of radius R and charge Q with uniform charge distribution using Equ 1 instead of Equ 2?

Equ 1:
ΔV = -∫Edotdx rather than

Equ 2:
ΔV = ∫dUelec potential

For Equ 1 I'm guessing that the equation would be of the form.....

ΔV = -∫∫Ecos(∏/2)dA = -∫∫EdA = -∫∫Edxdy
And
Limits of int (X-Direction): -√(R2 - y2) to √(R2 - y2)
Limits of int (Y-Direction): -√(R2 - x2) to √(R2 - x2)

And E would be the EField of a disk....

Last edited: Mar 15, 2013
2. Mar 16, 2013

### vela

Staff Emeritus
You can certainly integrate the E field to find the potential, but your attempt is wrong. It looks like you're confusing calculating the potential with calculating E.

To use equation 1, you need to have already found $\vec{E}$, and you'd calculate the line integral along any path C that runs from infinity to the point $\vec{r}$:
$$V(\vec{r}) = -\int_C \vec{E}\cdot d\vec{x}.$$ To find $\vec{E}$, you'd calculate the contribution from each piece of the disk to the electric field at some point in space, and then integrate over the entire disk.

To use equation 2, you take a similar approach as you do when calculating $\vec{E}$. You calculate the contribution from each piece of the disk to the electric potential at some point in space, and then integrate over the entire disk. Calculating electric potential is a bit easier than calculating the field because the potential is a scalar quantity. You don't have to worry about summing vectors like you do when calculating the electric field.

3. Mar 16, 2013

### Miike012

So equation 1 would be..

V = -∫Ediskdx and integrate from xf = distance from point to disk and xi = ∞. is that correct?

Last edited by a moderator: Mar 17, 2013
4. Mar 17, 2013

### vela

Staff Emeritus
Yes, basically.