Solve Charge Density (rod) Homework Problem

[jesuslovesu]

Never mind, i got it, whew

Homework Statement

http://img137.imageshack.us/img137/6250/chargqg0.th.jpg
The rod has a nonuniform charge density lambda = ax (a is a positive constant). Find electric potential at point A.

Homework Equations

V = kq/r
E = kqq/r^2
V = -$$\[ \int_a^b E dot dr\]$$

The Attempt at a Solution

I am pretty close to the answer, I'm just not quite there.
If I'm not mistaken it is something like:
$$\[ \int_d^{L+d} kax*r/r^2\,dr$$
but I'm not quite sure what to do with the 'x', from the answer, I know that it has to end up being the integral of 1/r, so x can't be r because it would end up being just the integral of dr

Maybe the limits are incorrect?

 

[Gianna_07]

Hi! I have a similar problem...

A rod of length L lies along the y-axis with its center at the origin. The rod has a nonuniform linear charge density lambda =a | y | , where a is a constant with the units {\rm C}/{\rm m}^{2} . Determine the constant a in terms of L and the rod's total charge Q.

I know that for uniform charge its lambda=L*Q and then integrate but what do I do in this case?

Thanks!

 

[Moetasim]

Any help would be greatly appreciated!
Great job on solving the problem! It can be tricky to work with nonuniform charge density, but it looks like you have the right idea. The limits are indeed incorrect - they should be from 0 to L, since that is the length of the rod. Also, you are correct that the 'x' term should not be included in the integral. It should simply be:
\[ \int_0^L kax/r\,dr \]
which simplifies to:
\[ ka\ln(L/d) \]
This is the correct potential at point A. Keep up the good work!