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Force of one distribution of charge on another

  • #1
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Homework Statement


I need help on solving this exercise :
We have a ring of radius = ##a## uniformly charged (total charge = ##Q##) and on its axis a segment ##OA## (length = ##a## also) of uniformly distributed positive electric charges with the charge density ##\lambda'## and of total charge = ##Q## (the same as the ring).

Zwcu3.png


I'm asked to determine the total force ##\vec{F}## that the ring's distribution is exerting on the segment's distribution.

Homework Equations


##\vec{F}=q.\vec{E}##

The Attempt at a Solution


First of all we have, for a point ##M## of ##OA## :
##\vec{F}_{Ring/M}=k\frac{Q.\lambda'dz}{a^2+z^2}\vec{u_z}## with ##k\frac{Q}{a^2+z^2}\vec{u_z}=## the electric field created at ##M##.
##\Rightarrow \vec{F}_{Ring/OA} = kQ\lambda'\int_0^a \frac{dz}{a^2+z^2}\vec{u_z} = kQ\lambda'\frac{\pi}{4a}\vec{u_z}##
Where have I gone wrong?
Thank you!
 

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Answers and Replies

  • #2
I believe that the result must have the following shape
$$\mathrm{Const}\cdot\int_0^\pi \int_0^a\frac{zdzd\varphi}{(z^2+a^2)^{3/2}}$$
where ##\varphi## is the angle on the circle
 
  • #3
haruspex
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Where have I gone wrong?
You need just the component of the field in the z direction.
 
  • #4
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You need just the component of the field in the z direction.
So multiply by a cosine? But shouldn't that be already taken into account since we're given the charge of the ring? I mean we're supposing that each point of the ring is creating a force.
 
  • #5
haruspex
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So multiply by a cosine? But shouldn't that be already taken into account since we're given the charge of the ring? I mean we're supposing that each point of the ring is creating a force.
But what direction is the force from a small element of the ring? Is it along the z axis?
 
  • #6
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But what direction is the force from a small element of the ring? Is it along the z axis?
Yes, considering symmetries
 
  • #7
haruspex
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Yes, considering symmetries
No, that's the overall result. What are the magnitude and direction of the force from a small element of the ring? What component of that is along the z axis?
 
  • #8
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No, that's the overall result. What are the magnitude and direction of the force from a small element of the ring? What component of that is along the z axis?
That would be ##\vec{F}=k\frac{dq_{ring}.dq_M}{a^2+z^2}cos\theta .\vec{u_z}##
 
  • #9
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Well the force of the whole ring on a point M would be $$\vec{F}_{Ring/M}=k\frac{Q.\lambda'dz}{a^2+z^2}\cos{\theta}.\vec{u_z}$$ then substitute the cosine with ##\frac{z}{\sqrt{a^2+z^2}}## and so $$\vec{F}_{Ring/OA} = kQ\lambda'\int_0^a \frac{z.dz}{(a^2+z^2)^{3/2}}\vec{u_z}$$
I guess that's it, thank you!
 
  • #10
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What made me rethink using the cosine here is the fact that we're given the total charge of the ring, in my mind I was literally picturing myself looking from "a charge on the ring" point of view and I was imagining lines from all the other charges, as well as the one i'm looking from, moving towards a point M and naturally the electric field at that point was in the ##\vec{u_z}## direction, so I thought maybe we ought to directly express the field as ##E=k\frac{Q}{a^2+z^2}\vec{u_z}##
 

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