Electric potential due to line of charge

[stickyrice581]

Homework Statement

http://img239.imageshack.us/img239/9112/54077471xz4.jpg

Homework Equations

http://img183.imageshack.us/img183/1450/clipboard01bt7.jpg

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?

:)

 

[alphysicist]

Hi stickyrice581,

Homework Statement

Homework Equations

http://img183.imageshack.us/img183/1450/clipboard01bt7.jpg

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?

:)

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?

 

[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

 

[stickyrice581]

Is there a formula I can use for my problem?

 

[alphysicist]

Is there a formula I can use for my problem?

king vitamin gave you the formula in his post:

<br /> V = k \int \frac{dq}{r}<br />
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.