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Field Due to Continuous Distribution of Charge

  • #1

Homework Statement



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


Homework Equations



E = [tex]\frac{1}{4\pi\epsilon}[/tex][tex]\int\frac{dq}{r^2}[/tex]
F = [tex]\int E dq[/tex]



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, [tex]l_{1}[/tex].

E = [tex]\frac{1}{4\pi\epsilon}[/tex][tex]\int\frac{dq}{r^2}[/tex]

Limits of integration being (0,

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

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

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

which is just [tex]\frac{\lambda_{1}}{4\pi\epsilon} * [/tex][tex]\frac{1}{a-l_{1}}[/tex]

and because a = d + [tex]l_{1}[/tex]

I get the E Field being E = [tex]\frac{\lambda_{1}}{4d\pi\epsilon}[/tex]

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.
 
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Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
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
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
rl.bhat
Homework Helper
4,433
7
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)
 

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