Two thin rods on the x-axis

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Homework Statement



Two thin rods of length L lie along the x-axis, one between x = a and x = a + L, and the other between
x = –a and x = –a – L. Each rod has positive charge Q distributed uniformly along its length.
a. Calculate the electric field as a function of x produced by the rod on the left hand side
(from –a – L to –a) at points along the positive x-axis.
b. Determine the magnitude of the force that one rod exerts on the other.


Homework Equations


E=∫[dq/4πε_naught*r^2] where dq=λdx for a line of charge


The Attempt at a Solution


Well since for part a, they only want the electric field in the positive x-axis I came up with:

E=∫from -a to -a-L of [λdx/4πε_naught*(-a-L+x)^2

giving me:

E=L/2a+L(2a+2L) when the integral is evaluated from -a to -a-L. - I'm aware I need to multiply my constants back in but for right now, I'm just worried about the integral.

Am I on the right track?
 

Answers and Replies

  • #2
SammyS
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Regarding your integral: E=∫from -a to -a-L of [λdx/4πε_naught*(-a-L+x)^2 ,

What does x represent?

You seem to have used x for two different things in the same expression.
 
  • #3
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x is suppose to represent where you are on the x axis.

-regarding the bump, sorry it won't happen again.
 
  • #4
SammyS
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x is suppose to represent where you are on the x axis.

-regarding the bump, sorry it won't happen again.
But you have (λ dx) in the integral, so in that sense, you're using x as the integration variable.

Use something like x0 (or some variable like u or s), for the place at which you are finding E. You then need to express the distance from x0 to the point x on the rod, over which you are integrating. --- r = x0 - x .
 
  • #5
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Sorry, I'm not following. The way I see it, when you take all the constants out you are just integrating dx/(-a-L+x)^2
 
  • #6
SammyS
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Yes, but that's not what you should be integrating.

(-a-L+x) is the distance between point x, and point (a + L) on the x axis.

What would the integral be if you were finding E at a fixed location, such as at x = 1.357 meter, on the x-axis ?
 
  • #7
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Okay, I think I know what you mean.

E=∫kdq/r^2 let dq=λdl and r =(a+x), the distance from the end of the rod to a +x value
E=λk/(a+x)^2∫dl -integrating from -a-L to -a


E=λk/(a+x)^2[(-a)-(-a-L)]

E=λkL/(a+x)^2

How's that look?

edit: I decided to use k rather than 1 over 4pi epsilon naught
 
  • #8
SammyS
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No. r should be the distance between the element of charge, dq, and x. So, r = x - l.
 

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