Find the electric field inside a circle

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  • #1
fluidistic
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Homework Statement


I set up myself to find the electric field inside a charged circle, but not in the center of the circle. Instead, in a point on a diameter, at a distance R/2 from the center, where R is the radius.
2. The attempt at a solution
I've tried to set up an integral summing the contribution of all the circle but without success. Is it a good idea? How would you proceed?
I know that the electric field must be 0, but as I'm not really convinced, I wanted to verify it.
Any help is appreciated.
dE=dQ/r².
[tex]Q=2 \pi R \lambda[/tex].
[tex]dl=xd\theta[/tex] where dl is a differential arc on the circle, [tex]\theta[/tex] is the angle measured from the horizontal diameter to the dl element and x is the distance between the point where I want to find the electric field and the dl element.
 

Answers and Replies

  • #2
AEM
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I would compute the field dE at your point P due to a bit of charge [itex] dq = \lambda rd\theta [/itex] and integrate around the circle. I would also look to see if there is any symmetry condition that makes solving the problem easier. (Note: this is my first take on how to approach it. There might be a better way.)
 
  • #3
fluidistic
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I would compute the field dE at your point P due to a bit of charge [itex] dq = \lambda rd\theta [/itex] and integrate around the circle. I would also look to see if there is any symmetry condition that makes solving the problem easier. (Note: this is my first take on how to approach it. There might be a better way.)
Ok perfect! That was the kind of reply I was hoping for. Thanks, I'll try to work something out.
 
  • #4
fluidistic
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I get [tex]d\vec E=\frac{Qd\theta}{2\pi Rr}\hat r[/tex] where R is the radius and r is the distance between P and dl.
I'm sure there's a [tex]cos \theta[/tex] that will appear from here, and integrating from 0 to [tex]2 \pi[/tex] I would eventually reach 0... but I don't know how to make the [tex]cos \theta[/tex] to appear. This is not new to me. I always struggle with this. :cry:
 
  • #5
AEM
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I get [tex]d\vec E=\frac{Qd\theta}{2\pi Rr}\hat r[/tex] where R is the radius and r is the distance between P and dl.
I'm sure there's a [tex]cos \theta[/tex] that will appear from here, and integrating from 0 to [tex]2 \pi[/tex] I would eventually reach 0... but I don't know how to make the [tex]cos \theta[/tex] to appear. This is not new to me. I always struggle with this. :cry:
Okay, so you got me hooked enough that I took another look at it. I drew a circle and drew a horizontal line I'll call the x-axis and a vertical y-axis. For each bit of charge [itex] \lambda r d\theta [/itex] there will be one symmetrically placed on the opposite side of the x axis. That means that the y components of the vector dE will cancel. That takes care of your [itex] cos(\theta) [/itex]. You need that to get the x component of dE. What concerns me is that th expression for r is going to be messy, but I'm going to have to sign off soon. I'm curious as to what you're going to come up with...
 
  • #6
fluidistic
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Okay, so you got me hooked enough that I took another look at it. I drew a circle and drew a horizontal line I'll call the x-axis and a vertical y-axis. For each bit of charge [itex] \lambda r d\theta [/itex] there will be one symmetrically placed on the opposite side of the x axis. That means that the y components of the vector dE will cancel. That takes care of your [itex] cos(\theta) [/itex]. You need that to get the x component of dE. What concerns me is that th expression for r is going to be messy, but I'm going to have to sign off soon. I'm curious as to what you're going to come up with...
You drew the same sketch than I. I also realized that the symmetry will automatically tell us that the E field in the y direction is null, but after all I wanted to prove it mathematically.
I'm also stuck as to write r... I'd have an "x" variable, but it doesn't seem easy to continue.
For example [tex]r=\sqrt{x^2+y^2}[/tex] or I could choose [tex]r=\frac{x}{\cos \theta}[/tex], though I don't know how to go any further. What limit to take in the integral for example.
[tex]\theta[/tex], 0 to [tex]2\pi[/tex]. But x...
I've no idea.
 
  • #7
rl.bhat
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If you start measuring x from the point R/2, using the properties of circle, you can write
y^2 = (R/2-x)*(3R/2+x)
You can take the integration form x = 0 to x = R/2 for smaller part and x=0 to x = 3R/2 for the larger part.
 
  • #8
fluidistic
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If you start measuring x from the point R/2, using the properties of circle, you can write
y^2 = (R/2-x)*(3R/2+x)
You can take the integration form x = 0 to x = R/2 for smaller part and x=0 to x = 3R/2 for the larger part.
I thank you for your help and time. I don't understand how you reached this. Is it be simple Pythagorean theorem? I only see y²=r²-x² or y=r sin (theta).
Time for me to go to sleep, it's past 2 am and if I don't sleep now I won't sleep the night before the exam, ahah.
 
  • #9
rl.bhat
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The relation is true for any intersecting diameter and a cord which is perpendicular to the diameter. Cord bisects the diameter.
Let AB be the diameter and PQ is the cord. Let O be the point of intersection.
Triangle OAP and OBQ are similar. So the sides are proportional.
Hence OP/OB = OA/OQ Or
OA*OB = OP*OQ. But OP = OQ. So
OA*OB = OP^2.
 

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