# Potential due to a continuous charge distribution.

1. Mar 10, 2008

### Ajwrighter

1. A nonconducting rod of length L = 6cm and uniform linear charge density A = +3.68pC/M . Take V = 0 at infinity. What is V at point P at distance d = 8.0cm along the rod's perpendicular bisector?

2. V = S E * ds One half of the rod = L/2 1/4piEo = 9x10^9
R = sqrt((L/2)^2 + (D^2)) = 0.08544m

3. I've attempted this solution many times but here is the more recent.
(9x10^9) [(((3.68 x 10^-12)/0.08544 ) * (.03)*(2) ] = .0232584V

My reasoning for this method. Take the Charge density A and divide it by the Range of both sides, then multiply by the length of each side which equals .03 and then multiply it by twice to obtain V from both sides (since both sides are identical.) The problem is the answer in the back gives 24.3mV = .0243V. The method I am using, is it just a close proximity, a lucky guess? Or can my method be reproduced on similar problems to obtain a close proximity? In either case, if my method is wrong what do I need to change?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 10, 2008

### pam

You have to integrate over the length of the rod.
The calculation is the same as that in many elementary texts for an infinite rod,
but with finite limits on the integral.

3. Mar 10, 2008

### Ajwrighter

on one of my previous methods when I integrated I get 0.02484

4. Mar 10, 2008

### Ajwrighter

I just tried integrating another way and got 0.02358.