Finding the force from two charged rods one beside the other.

  • Thread starter Eats Dirt
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  • #1
Eats Dirt
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Homework Statement



The problem is to find the electric force between two rods one beside the other. a charge of Q is distributed over the rods uniformly each of length 2x. The centre of each rod is d from the others centre.




-----Rod 1------- ------Rod 2-------



Homework Equations



[tex]F= \frac{qq}{Kr^2}[/tex]

[tex]Efield= \frac{q}{Kr^2}[/tex]



The Attempt at a Solution



I know I am supposed to find the electric field over the whole distance then integrate the the length of one rod (I think) but I don't understand at all why this work

I would set the electric field up by

dq = Q/2x (Charge density) * dx

E= dq/(Kr^2) I think. then I am not even sure if I did that right and I think I'm supposed to integrate this over the length of 1 of the rods but I don't understand how that gives you force. Any help would be appreciated thanks!
 
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Answers and Replies

  • #2
voko
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You need to find the electric field of a rod at an arbitrary location along its axis (outside itself). Then you need to consider how that field acts on another rod. You can do that by considering how it acts on each particle of the other rod - what is the net force then?
 
  • #3
Eats Dirt
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You need to find the electric field of a rod at an arbitrary location along its axis (outside itself). Then you need to consider how that field acts on another rod. You can do that by considering how it acts on each particle of the other rod - what is the net force then?

ok at an arbitrary distance outside of the first rod would would be "d" then integrate with respect for dx for the length of the first rod. What I don't get is how to do the second integral, what do we integrate with respect to? The distance "d" to the second rod? or do we just keep d and do the integral with respect to the length of the second rod?
 
  • #4
voko
6,054
391
At any given point of the second rod, the electric fields of the first rod (found with the first integral) exerts a force on the second rod.
 

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