Potential Difference Near a Charged Rod

[lightupshoes8]

Homework Statement

A long rod of length L = 7.8 m carries a uniform charge Q = -6.5e-08 coulombs.
Calculate the potential difference VC - VA. The distance c is 4 cm, the distance b is 11 cm, and the distance a is 4 cm. Only part of the rod is shown in the diagram (it is longer than shown).

Homework Equations

Erod = (9e9)*\frac{2Q/L}{r}
\DeltaV = - \int E dl

The Attempt at a Solution

I figured that since A is the same distance horizontally as B from the rod, there would be no change in potential since its not acting in that direction. So I was just going to find Vc-Vb

I set up my integral as follows:

V = - (9e9)*\frac{2Q}{L}\int \frac{dr}{r}
Integral going from upper to lower: B to C

I just don't think I'm headed in the right direction. It seems like the answer is not going to be what I want it to be.

 

[LowlyPion]

I figured that since A is the same distance horizontally as B from the rod, there would be no change in potential since its not acting in that direction. So I was just going to find Vc-Vb

That's right.

I set up my integral as follows:

V = - (9e9)*\frac{2Q}{L}\int \frac{dr}{r}
Integral going from upper to lower: B to C

That looks ok too.