Went back to the field of charge around a ring problem

In summary, the electric field at a point P that lies on the axis of the ring at a distance x from its Center is given by Coulomb's law.
  • #1
seanm1
3
0
I was going back to a previous chapter to study for my finals, and came across the field of charge around a ring problem. Basically, it's designed to show how using a special case can make a problem easier.

This was one of those equations I just kind of had to memorize and use. I'm the kind of person that can't learn from shortcuts easily and has to derive things from scratch to really understand them. So I tried to derive the equation by myself, and then looked at the book derivation, and everything promptly stopped making sense to me.

Homework Statement


A ring-shaped conductor with radius a carries a total charge Q uniformly distributed around it. Find the electric field at a point P that lies on the axis of the ring at a distance x from its Center.

(Note: the figure here shows a ring with the entire "O" visible on the page, and a line originating at the center of the ring and leading to the right forming the x axis, and a point along the x-axis well past the radius of the ring marked P, so that every point on the ring and P are all in the same plane).

Homework Equations


Coulomb's law, E = (kq/r^2) kg m A^-1 s^-3 \hat r
The ring is centered around the origin, the ring's radius a is given, the distance from the origin to the point analyzed on the X axis P is given, the total charge Q distributed evenly through the whole ring is given.

The Attempt at a Solution


On my own, I tried to set up the integral. I new later that the upper part of the ring and the lower part of the ring would mean that there would be no translation force along the y axis. But I'm still not quite sure how to get to that part.

First, I cut the ring into several differential angles. So far it's just [tex]\int_0^{2 \pi} d \theta[/tex] .

I reasoned that every differential angle would have a resultant charge. At any angle theta, the portion of the charge contained in the sweep of the ring inside that angle would be theta/2pi * Q. So the partial charge dQ would be Q d \theta/2pi.

This is where I got into trouble, because my reasoning differed from the book's at this point. This is also where I stopped being able to follow the book's explanation. I reasoned that the lateral distance from a point on the ring at an angle theta to the point P would be

P - a cos \theta

and the vertical distance would simply be

a sin \theta

So the radius squared would be, simply,

[tex](P - a cos \theta)^2 + a^2 sin^2 \theta[/tex]

That simplifies through trig identities like so:

[tex]P^2 - 2 a P cos \theta + a^2 cos^2 \theta + a^2 sin^2 \theta[/tex]

[tex]P^2 - 2 a P cos \theta + a^2(1 - sin^2 \theta) + a^2 sin^2 \theta[/tex]

[tex]P^2 - 2 a P cos \theta + a^2[/tex]

Which makes my integral,

[tex]\int_0^{2 \pi} k \frac {Q d \theta}{(P^2 - 2 a P cos \theta + a^2) 2 \pi}[/tex]

or

[tex]\frac {kQ}{2 \pi} \int_0^{2 \pi} \frac {d \theta}{(P^2 - 2 a P cos \theta + a^2)}[/tex]

Well that was simple enough to look up (or try to solve with partial fractions)! Unfortunately, it was also simple enough to get me into trouble, the result was

[tex]\frac {kQ}{\pi (r^2 + P^2)} [ arctan(\frac {(P + r)(tan (\theta/2))}{P - r}]_0^{2 \pi}[/tex]

Of course, thanks to that tan \theta/2, that evaluates to zero, so rather than go any further (because I suspect I messed up my math somewhere along the line), I figured I'd ask the following two questions:

1. IS there a general case for this problem (that doesn't involve anything nasty/unsolveable like eliptic integrals, that works for any point in 2-space outside the ring), and

2. How are you supposed to derive this answer (using the shortcut; this is a solved problem I just couldn't follow the explanation I had).

Fake edit: After typing all this out, I came to realize that the book was actually using the angle between the axis and the hypotenuse (which would keep x static) rather than integrating over the ring angles. That does make the textbook explanation make a LOT more sense, but I'm still interested in seeing if there's a way to do it without taking into account everything in the y-axis canceling out.

I'm a bit too frustrated with the dead end I was going down using the ring angles instead of the angle my textbook used to try now, but it would seem to me that the general case shouldn't be too difficult to find here.
 
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  • #2
Hi,

I'm finding a contradiction between the problem statement and your description of the figure:

seanm1 said:

Homework Statement


A ring-shaped conductor with radius a carries a total charge Q uniformly distributed around it. Find the electric field at a point P that lies on the axis of the ring at a distance x from its Center.
Let me point out that "the axis of the ring" is perpendicular to the plane of the ring, and intersects that plane at only one point, at the center of the ring.

(Note: the figure here shows a ring with the entire "O" visible on the page, and a line originating at the center of the ring and leading to the right forming the x axis, and a point along the x-axis well past the radius of the ring marked P, so that every point on the ring and P are all in the same plane).
This description of the x-axis, as a line lying in the plane of the ring, is different than the previous description that it is the axis of the ring.

Can you clarify this discrepancy? Perhaps it would be good to look at the figure again.

Also, it appears that the integral you set up was adding up the magnitude of the electric field from each charge element dQ. But you must use vector addition when adding up electric fields -- so set up an integral for the x-component of the electric field instead.
 
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  • #3
Thanks a lot! On both accounts, I was both getting the axis wrong (if the ring is in the xz plane I should use the y axis, which would actually make "the ring axis" make more sense now that I think about it), and should have been using vectors.

I always end up missing the really obvious stuff (and it usually bites me hard at test time). It was hiding right in front of me, where I would never have seen it. I'm just grateful that you were able to notice what I was getting wrong right off despite my convoluted explanation.

I'll come back in a little bit to post my reworked solution so it's there for other people. Thanks again!
 
  • #4
seanm1 said:
... Thanks again!
You're welcome :smile:
 
  • #5


I can understand your frustration with trying to derive the equation for the electric field around a ring of charge. It can be a complex problem, and sometimes it is easier to just memorize the equation. However, I would encourage you to keep trying to understand the derivation, as it will deepen your understanding of the concept and help you in future problem-solving.

To answer your first question, yes, there is a general case for finding the electric field around a ring of charge. It involves using the concept of infinitesimal charge elements and integrating over the entire ring. This is similar to what you were attempting to do in your own derivation, but it requires a bit more mathematical manipulation.

To answer your second question, there are multiple ways to derive the equation for the electric field around a ring of charge. The method your textbook used is one way, but as you discovered, it can be confusing and difficult to follow. Another approach is to use Gauss's law, which simplifies the calculation by considering the symmetry of the problem. There are also other methods, such as using the concept of potential and integrating over the ring.

Ultimately, the best approach is the one that makes the most sense to you and helps you understand the concept. If you are struggling with one method, don't be afraid to try a different approach or seek help from a classmate or teacher. Remember, understanding the concept is more important than just memorizing the equation. Keep practicing and you will eventually master it.
 

1. What is meant by "field of charge around a ring problem"?

The "field of charge around a ring problem" refers to a theoretical scenario in which a charged ring or circular object is placed in an electric field. This problem is commonly studied in the field of electrostatics, which deals with the behavior and interaction of electric charges.

2. How is the field of charge around a ring problem solved?

The field of charge around a ring problem is typically solved using mathematical equations and principles from electrostatics, such as Gauss's Law or Coulomb's Law. These equations help to calculate the electric field and potential at different points around the charged ring.

3. What factors affect the field of charge around a ring problem?

The strength of the electric field around a charged ring is affected by several factors, including the amount of charge on the ring, the distance from the ring, and the orientation of the ring relative to the electric field. The presence of other nearby charged objects can also have an impact on the field.

4. Can the field of charge around a ring problem be applied to real-world situations?

Yes, the principles and equations used to solve the field of charge around a ring problem can be applied to real-world situations and have practical applications in fields such as telecommunications, electronics, and energy systems.

5. Are there any limitations to the field of charge around a ring problem?

Like any theoretical problem, the field of charge around a ring problem has its limitations. It assumes ideal conditions, such as a perfectly symmetrical and uniform charge distribution, and does not take into account factors such as the motion of the charged ring or the effects of magnetic fields. Real-world situations may require more complex calculations and considerations.

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