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# Proof: Field from infinite plate (part 1)

Advanced proof of the formula for the electric field generated by a uniformly charged, infinite plate. Created by Sal Khan.

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• this is a really dumb question, but the parallel plates that aren't infinite are just an approximation of the infinite when charge and length of the plate are big enough compared to the distance right?
• Hi.

You are sort of correct. Khan actually mentions that at the end of the next video in this series: that for a finite plate where you are far away from the edge, this theory holds, so yes, you are correct. However, just remember that right near the edge, things fall apart.

Most theories are really good approximations; the only real values come when you test experimentally.

This isn't really a dumb question; I pondered in once myself, and most people will.
• At , why is the area of the ring the circumference times dr? If dr were much larger (call it w), wouldn't the area be pi((r+w)^2-r^2) ? Like the difference between two circles.
• You are actually correct. If dr was larger, than it would be pi((r+w)^2-r^2) for the area. However, this is not the place to use that. We are assuming an infinitely small width between the two circles, we just assume that there is a central circle, right in the middle of the two circles, and we extend a length dr for the width. Because of the size, this average value is actually really accurate, so we stick to it, and that makes the math much more simple.

I hope that is not confusing; one of my teachers spent half a class teaching us why that is so, but his explanation was much better than mine.
• Around , Sal says that the force generated by the ring is = kqq/d^2. However, how is the entire charge of the ring enclosed to one point? More specifically, how come he is treating the entire charge of the ring as enclosed to one point? Is he allowed to do that? Thanks.
• Since the ring is of uniform charge, and the charge density is sigma then Q (<-Charge of ring) = sigma (<-charge density) * (2 * pi * r * dr) (<-area of the ring). He is correct in doing this, but only in the case of a surface of uniform charge (The charge at every point of the surface is equal) and with the test charge (q) at the center of the ring so that all of the horizontal components of the electrostatic forces cancel, leaving only the vertical components.

Since forces are vector quantities and in the same direction, we can treat them as if they were all directly under the test charge and them sum them up to receive the total net force on the test charge from the ring.

Now if the ring was not of uniform charge and/or the test charge was not at the center of the ring, then we would have a much more difficult problem.

I hope this clears things up for you. ^^
• At why is the width of the circle 'dr'; why does it share the same variable as the radius? Why not 'dx' or 'dw'? What allows us to use 'r' both times? Does it have to do with mstocksl@hotmail.com's question? Thanks!
• because dr is just a small change in radius
since he's talking a bout a circle with radius r it seems fitting to use dr
and because in the next video he's going to integrate across r
if he used dw he would have to rename r to w to be consistent
(1 vote)
• Doesn't this video assume that the magnetic field will repel the particle, instead of attracting it?
• Yes it does. It was mentioned early in the video that the the chanrge on the plate is positive and the particle is positive as well so the force would be repulsive.
• Why does Sal take an infinite plane? I find it quite hard to imagine.
• If you pick a plane of finite size, you would have to compute the net effect of whatever you are studying at that boundary and include it in your final answer. In much of physics, an effect (gravitational potential or electric field for examples) are defined as zero at infinity, so you can ignore that 'boundary' value.

So think of it as 'choose a plane whose size is large enough that we can use infinity as the 'far' distance (for example, we could calculate (or integrate) from zero radius to infinity). If the effect has dropped to zero at infinity, that term becomes simpler to calculate - or may drop out altogether.
• At ~, why didn't we take πr^2, since there are charged particles everywhere on the ring?
• We have to consider the point charges on the ring. So the total charge isn't the area because the "ring" is assumed to be hollow. So the charge is :-
(circumference) * (thickness) * (charge density)

Watch the next part and Sal will integrate this equation and obtain a general formula for this situation.
• He saying i need to learn about calculus first, can anybody tell me from where should i start learning it, i mean there are 3 different calculus topics ;Differential, multivariable, integral.Or any other type?
• For this specific video, you'd want to learn basic integration techniques, though learning most of differential and integral calculus (single-variable) would be a good idea so as to establish a thorough knowledgebase.