Electric field of a semi circle ring

  • #1
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Homework Statement



5jXScWt.jpg


Suppose the charge Q on the ring of Fig. 28 was all distributed uniformly on only the upper half of the ring, and no charge was on the lower half. Determine the electric field [itex]\vec {E}[/itex] at P.

Homework Equations


[/B]
[tex] E = k\frac {q} {r^2} [/tex]

The Attempt at a Solution


[/B]
Starting with the charge density,

[tex]\lambda = \frac{Q}{\pi a} = \frac{dq}{dl}[/tex]
electric field for dq
[tex] dE = k\frac{dq}{x^2+a^2}[/tex]
Let's find its x component
[tex]dE_x = dEcos\theta = k\frac{dq}{x^2+a^2}cos\theta = kx\frac{dq}{(x^2+a^2)^{3/2}}[/tex]
[tex]dE_x = \frac{k\lambda x}{(x^2+a^2)^{3/2}}dl[/tex]

Wıth the limits being [itex]0[/itex] and [itex]\pi a[/itex]

[tex]E_x = \frac{k\lambda x \pi a}{(x^2+a^2)^{3/2}} = \frac{kqx}{(x^2+a^2)^{3/2}}[/tex]
This is correct according to my textbook.
However, I end up with a wrong y component when I apply the same logic for the vertical.
[tex]E_y = -\frac{k\lambda a \pi a}{(x^2+a^2)^{3/2}} = -\frac{kqa}{(x^2+a^2)^{3/2}}[/tex]

But the correct answer for the vertical is
[tex]E_y = -\frac{kq2a/\pi}{(x^2+a^2)^{3/2}}[/tex]

Having searched the internet for this question, I found that the solution for the y axis includes an arc length element, which is something like this.
[tex]\lambda = \frac {dq} {a d\theta}[/tex]
Then I suppose, it integrates from [itex]0[/itex] to [itex]\pi[/itex] for a semi circle.

So my question is, how does that differ if one uses [itex]d\theta[/itex] or [itex]dl[/itex] for integration? Also, why did I get the x component correct, even though the fact that I used [itex]dl[/itex]?
 

Answers and Replies

  • #2
haruspex
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So my question is, how does that differ if one uses [itex]d\theta[/itex] or [itex]dl[/itex] for integration? Also, why did I get the x component correct, even though the fact that I used [itex]dl[/itex]?
For the x component, the contribution of each element was independent of theta. Not so for the y component.
 
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  • #3
TSny
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You didn't show the details of your work for the y component.

Note that using sinθ instead of cosθ will give you the component of ##d\vec{E}## that is perpendicular to the x axis. But this perpendicular component is not parallel to the y axis (in general).
 
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  • #4
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You didn't show the details of your work for the y component.

Note that using sinθ instead of cosθ will give you the component of ##d\vec{E}## that is perpendicular to the x axis. But this perpendicular component is not parallel to the y axis (in general).
It is much like the work I've done for the x component, but let me write it any way.

[tex]dE_y = dEsin\theta = k\frac{dq}{x^2+a^2}sin\theta = ka\frac{dq}{(x^2+a^2)^{3/2}}[/tex]
[tex]dE_y = \frac{k\lambda a}{(x^2+a^2)^{3/2}}dl[/tex]
[itex]0[/itex] to [itex]\pi[/itex]
[tex]E_y = \frac{k\lambda a \pi a}{(x^2+a^2)^{3/2}} = \frac{kqa}{(x^2+a^2)^{3/2}}[/tex]

As you say, I'm probably missing something geometrical, but I can't see how dE*sin(theta) is not parallel to the y axis in this particular situation. If its cosine component is parallel to the x axis, shouldn't its sine component be parallel to the y axis?
 
  • #5
TSny
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As you say, I'm probably missing something geometrical, but I can't see how dE*sin(theta) is not parallel to the y axis in this particular situation.
Consider an infinitesimal charge element located at one end of the semicircular ring. In what direction does the perpendicular component dE*sin(theta) point? (I'm imagining the lower half of the ring chopped off and removed so the semicircle that is left has two ends.)
 
  • #6
haruspex
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If its cosine component is parallel to the x axis, shouldn't its sine component be parallel to the y axis?
There are three dimensions here. If the cosine gives the component parallel to the x axis then the sine gives the component orthogonal to the x axis, but that does not make it always parallel to the y axis.
 
  • #7
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Consider an infinitesimal charge element located at one end of the semicircular ring. In what direction does the perpendicular component dE*sin(theta) point? (I'm imagining the lower half of the ring chopped off and removed so the semicircle that is left has two ends.)
I think it doesn't have a y component because the ends of the circle is at the same height with point P, so I believe there is no y since there is no elevation.
 
  • #8
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There are three dimensions here. If the cosine gives the component parallel to the x axis then the sine gives the component orthogonal to the x axis, but that does not make it always parallel to the y axis.
Yes, now I got it. I've been playing with my hands for the last 10 minutes, making a circle with one and trying to understand what is going on. I see that all sine components cannot be added together because they do not point the same direction. Am I right?
 
  • #9
TSny
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Yes. Good.
 
  • #10
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Yes. Good.
I understand that we are in a 3d coordinate system, and sine component is perpendicular to x axis. Also, that perpendicular component is not always parallel to the y axis. So how can I proceed to the y components knowing these?
Thank you so much for your help by the way.
 
  • #11
TSny
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Pick an element of charge and imagine a line drawn from the origin to the element. This line will make some angle φ to the y axis. Imagine a plane that contains this line and the x axis. Can you see that the electric field produced by the element of charge will lie in this plane? So, the component of the electric field that is perpendicular to the x axis will also lie in this plane. See if that helps in finding the y component of the field.
 
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  • #12
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Pick an element of charge and imagine a line drawn from the origin to the element. This line will make some angle φ to the y axis. Imagine a plane that contains this line and the x axis. Can you see that the electric field produced by the element of charge will lie in this plane? So, the component of the electric field that is perpendicular to the x axis will also lie in this plane. See if that helps in finding the y component of the field.
cLJeY0T.jpg


If my setup looks correct, then I would like to ask how to find a relationship between beta and theta.
 
  • #13
TSny
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You won't need a relationship between those two angles. Notice that θ is a fixed angle while β varies to cover all the charge elements. Your diagram looks good. How would you project ##dE_\perp## onto the the vertical dotted line?
 
  • #14
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You won't need a relationship between those two angles. Notice that θ is a fixed angle while β varies to cover all the charge elements. Your diagram looks good. How would you project ##dE_\perp## onto the the vertical dotted line?
Multiplying it with [itex]cos \beta[/itex]?
If I am right, then I need limits for the integral. Would it be 2*[from pi/2 to zero], or 2*[zero to pi/2]?
 
  • #15
TSny
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I would use twice the integral from 0 to pi/2.
 
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  • #16
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I would use twice the integral from 0 to pi/2.
Thank you so much, I've managed to get the correct answer.
Just one more thing. Should the final result's sign always obey the direction of the vector?
 
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  • #17
TSny
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Yes. If you are asked for Ey, then the answer should be negative. If you are asked for the magnitude of the y component, then that generally means just the absolute value.
 

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