# Homework Help: Field Due to Continuous Distribution of Charge

1. Dec 13, 2009

### discordplus

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

Coulomb force between line charges: a rod of length l1 with line charge density λ1 and a rod of length l2 with line charge density λ2 lie on the x axis. Their ends are separated by a distance D as shown in the figure.

(a) What is the force F between these charges?

diagram: http://ocw.mit.edu/NR/rdonlyres/Physics/8-022Fall-2004/3A772032-6B74-4D2D-A550-8F0ECFECEDBC/0/pset1.pdf [Broken]

#7

2. Relevant equations

E = $$\frac{1}{4\pi\epsilon}$$$$\int\frac{dq}{r^2}$$
F = $$\int E dq$$

3. The attempt at a solution
So, first I decided to find the field at a point a distance D from the end of line 1. Using the standard x coordinate system, I placed line 1 such that its endpoints are 0, $$l_{1}$$.

E = $$\frac{1}{4\pi\epsilon}$$$$\int\frac{dq}{r^2}$$

Limits of integration being (0,

Using this and dq = $$dl_{1}$$$$\lambda_{1}$$, all I need to do is find a function for r in terms of l, which is the distance from 0. Which would be ($$l_{1}$$ + D) - l.

I renamed ($$l_{1}$$ + D) as the variable a to make the integration simpler. So now I have:

E = $$\frac{\lambda_{1}}{4\pi\epsilon}$$$$\int \frac{dl}{(a - l)^2}$$

which is just $$\frac{\lambda_{1}}{4\pi\epsilon} *$$$$\frac{1}{a-l_{1}}$$

and because a = d + $$l_{1}$$

I get the E Field being E = $$\frac{\lambda_{1}}{4d\pi\epsilon}$$

Is this correct so far? Clearly my success on the second part depends on that because all I have to do is just integrate the field over the infinitesimal segments of charge over the second line's length yes? And to find that distance d as a function of l it's just ( l - length 1), where l is the distance from the 0 point. I'm just kind of shaky on the first part, finding the field, that's all.

Last edited by a moderator: May 4, 2017
2. Dec 13, 2009

### rl.bhat

In the expression of E at the starting point of l2, the distance l1 must appear which is missing in your expression.
Try this one.
dE = k*lambda1*dx/(D+x)^2.
Integrate it from x = 0 to x = l1 and find E.

3. Dec 13, 2009

### discordplus

Wait, is x the distance from l1? Because then D + x is the distance from some point on the line to P2 which is what i want. Because my expression actually has d in it? It's just "a" is in terms of D, and i defined my distance variable as from x = 0. I'm not sure, it looks like x+d only works if x is the distance from l1. Or am I misunderstanding something here?

4. Dec 13, 2009

### rl.bhat

You want to find the field due to rod1 at the starting point of rod 2.
So E = k*lambda1*[-(1/(x+D)]. Find the value of E taking the limits from x = D to x= (l1 +D)