# Homework Help: Electric potential due to line of charge

1. Oct 24, 2008

### stickyrice581

1. The problem statement, all variables and given/known data

http://img239.imageshack.us/img239/9112/54077471xz4.jpg [Broken]

2. Relevant equations

http://img183.imageshack.us/img183/1450/clipboard01bt7.jpg [Broken]

3. The attempt at a solution

The online chapter gave me the above equation for the potential due a continuous charge. I converted pico coulomb to coulomb and cm to m. I also did charge times length for lambda. I got .0644 Volts and the site says its wrong. So then I tried charge divided by length for lambda. I got 3.28 Volts and it's also wrong. Any ideas?

:)

Last edited by a moderator: May 3, 2017
2. Oct 25, 2008

### alphysicist

Hi stickyrice581,

The formula that you have does not apply to this situation. That formula was calculated for a uniform charge density. However, in your problem the charge density is different at different parts of the rod.

So you'll need to calculate an analogous formula for your case. What do you get?

Last edited by a moderator: May 3, 2017
3. Oct 25, 2008

### king vitamin

I think you have to integrate the charge density for Q. It actually greatly simplifies the integral.

V = k*int(dq/x) from .05m to .19m

Now the trick is rewriting the dq in terms of dx

4. Oct 25, 2008

### stickyrice581

Is there a formula I can use for my problem?

5. Oct 25, 2008

### alphysicist

king vitamin gave you the formula in his post:

$$V = k \int \frac{dq}{r}$$
where the integral is taken over the charge distribution.

If you look in your book where you found the formula you have in your first post, you should see how they use this integral for the uniformly charged rod. Just follow the same type of procedure, using the fact that this rod is non-uniformly charged.