# Homework Help: What is the total force on a line charge due to another? Hw question

1. May 13, 2010

### ProtonHarvest

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

Two line charges of the same length L are parallel to each other and located in the xy plane. They each have the same linear charge density λ=constant. Find the total force on II due
to I.

2. Relevant equations
F =QE
Q = $$\lambda$$L
E = ? This is where I'm stuck.

3. The attempt at a solution
This got me a 0 points on the homework so I know its very wrong. I really just need some clues as to how to set up the integral, or if this is just some kind of trick question, what makes it tricky. I've got tunnel vision at this point so any help will be greatly appreciated.

$$d\vec{E} = \frac{\lambda dl \hat{r}}{4\pi\epsilon a^{2}}$$

$$\vec{E} = \frac{1}{4\pi\epsilon}\int\frac{\lambda dl\hat{r}}{a^{2}}$$

=$$\frac{L\lambda}{a^{2}4\pi\epsilon}$$

$$\vec{F} = \frac{(L\lambda)^{2}}{a^{2}4\pi\epsilon}$$

2. May 13, 2010

### nickjer

Use Gauss's law to find the electric field generated by an infinite line charge.

EDIT: Nevermind, just read they weren't infinite.

You will have 2 integrals. One for the electric field from wire 1 (and your setup for the E-field from wire doesn't look right at all, you need some z-dependence in it). You will also need a 2nd that runs over the 2nd wire summing up $\lambda E_x(y_2) dy_2$ to get the net force in the x-direction. This is because the electric field isn't constant over the length of the wire, so you can't just say F = E*Q.

Last edited: May 13, 2010
3. May 13, 2010

### ProtonHarvest

Wouldn't the radial component be $$E_{x}$$, & depend only on y and x, since this is a strictly 2-dimensional problem?

Using the hyperphysics site and the setup you see in the picture, I got:

$$dE_{x} = \frac{\lambda xdy}{4\pi\epsilon r^{2}r}$$

$$E_{x} = \frac{\lambda}{x4\pi\epsilon}\frac{L}{\sqrt{x^{2}+L^{2}}}$$

Aside from that, thank you so much for the input, you broke my tunnel vision! I can see what to do now in principle and its very helpful.