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seanm1
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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.
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).
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.
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 \int_0^{2 \pi} d \theta .
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,
(P - a cos \theta)^2 + a^2 sin^2 \theta
That simplifies through trig identities like so:
P^2 - 2 a P cos \theta + a^2 cos^2 \theta + a^2 sin^2 \theta
P^2 - 2 a P cos \theta + a^2(1 - sin^2 \theta) + a^2 sin^2 \theta
P^2 - 2 a P cos \theta + a^2
Which makes my integral,
\int_0^{2 \pi} k \frac {Q d \theta}{(P^2 - 2 a P cos \theta + a^2) 2 \pi}
or
\frac {kQ}{2 \pi} \int_0^{2 \pi} \frac {d \theta}{(P^2 - 2 a P cos \theta + a^2)}
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
\frac {kQ}{\pi (r^2 + P^2)} [ arctan(\frac {(P + r)(tan (\theta/2))}{P - r}]_0^{2 \pi}
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.
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 \int_0^{2 \pi} d \theta .
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,
(P - a cos \theta)^2 + a^2 sin^2 \theta
That simplifies through trig identities like so:
P^2 - 2 a P cos \theta + a^2 cos^2 \theta + a^2 sin^2 \theta
P^2 - 2 a P cos \theta + a^2(1 - sin^2 \theta) + a^2 sin^2 \theta
P^2 - 2 a P cos \theta + a^2
Which makes my integral,
\int_0^{2 \pi} k \frac {Q d \theta}{(P^2 - 2 a P cos \theta + a^2) 2 \pi}
or
\frac {kQ}{2 \pi} \int_0^{2 \pi} \frac {d \theta}{(P^2 - 2 a P cos \theta + a^2)}
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
\frac {kQ}{\pi (r^2 + P^2)} [ arctan(\frac {(P + r)(tan (\theta/2))}{P - r}]_0^{2 \pi}
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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