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## Electrical engineering

### Unit 5: Lesson 2

Fields, potential, and voltage

# Proof: Field from infinite plate (part 2)

We see that the infinite, uniformly charged plate generates a constant electric field (independent of the height above the plate). Created by Sal Khan.

## Want to join the conversation?

• At , is he saying he can put a positive test charge at different heights h1, h2, h3, etc where h1 != h2 != h3, and that the net force on that charge due to the electric field of the plate will be the same in all cases? My understanding was that the further the positive test charge is from the plate, the lower the magnitude of the force on that charge due to the electric field will be; but I am not hearing him say that. I'm confused. Will someone clarify?
• It's because the plate is infinite. The forces applied to the q charge, as a result from horizontally distant charges on the plate, are close to horizontal, so their horizontal components are big(but cancel out) and their vertical components are insignificant. The higher q is, the smaller the horizontal component gets as the vertical grows(the angle θ grows), the more impact distant charges have.
• Sal says that no matter at what height the test point is the strength of the field remains same. If this is so then the force exerted by the field should remain constant. But as far as i know the magnitude of force keeps changing as we keep on changing the distance(i.e height in this case). Please help me. I'm so confused.
• Imagine a bunch of strings connecting each point on the plate to your test point. When you're really close to the plate, the repulsive force is extremely strong since you're very near the plate points directly underneath you, but you're at such a shallow angle to the rest of the plate's points that their y-components make relatively little impact. The further away those points are, the shallower the angle relative to you. As you move a little bit away, the distance between you and the points directly below you increases, but the angle between you and every other point on the plate gets steeper. The y-components of those now-steeper angles gets larger, and they compensate for the increased distance. The further away you move from the plate, the more you increase the influence of ever more-distant points on the plate.
• Why do you use integrals?
Isn't it more convenient to use r as the "thickness" of the ring and then define r as infinite?(to get an infinite wide circle)
K*h*2*pie*sigma*infinity²/(h²+infinity²)^(3/2)
• You confusion lies in a misunderstanding of the point of the ring. The ring (not a circle, but rather the circle's circumference) is the collection of points at some horizontal distance r away from the point charge. The net electric field that he calculates from the ring is the sum of the electric force of every point on that ring, not every point that lies within that circle, as you seem to imply. So he uses calculus to add up the electric force of every possible ring.

But that leaves a part of your question still unanswered. Why didn't he just calculate the net electric force generated by a circle (not ring) with infinite radius? Because to find the electric force generated by any circle on the plate, he'd have to do the same integral (the sum of all enclosed rings). So really, he's doing exactly what you said. He's finding the net charge by drawing an infinitely large circle with rings because you can't really calculate it any other way.
• Does this mean if you had 2 infinite and identically charged plates, a test charge would feel an equal force from either plate even if it was 1mm from one plate and 100km from the other? And if this can be extended to gravity does that mean you would be weightless regardless of your distance from equally dense infinite plates?
• That is correct the forces from the two plates would cancel each other.

Another interesting solution is the force on a charged particle in a charged sphere or a mass inside a massive sphere. The net force is zero.
• How did he decide that the charge of the ring, Qr=2*(π)r*d*r(sigma)? I know that 2*(pr)*r is circumference. How did he figure d*r=width? What is d, distance? Why doesn't he just use the area of the circle Area = π × r^2 equation to solve for Qr? Please let me know if you are able to explain and where I can figure this out. Thank you! :-)
• I didn't watch the video, but I'm guessing he is integrating the area of concentric rings to derive the E field from an infinite plate. In that case the area of a ring with radius r and width dr would be the length of the ring (ie. the circumference) times the width. So the incremental ring area is
dA = 2*π*r*dr
where the incremental charge on the ring is the ring area times charge per unit area (σ), so dQ = σ*dA
and the total charge Q is the integral of σ*dA from r = 0 to infinity.
• The field is infinitely large; therefore, wouldn't the force applied to the test charge be infinitely strong as well?
• adding infinite amount of quantities doesn't always result in infinity, think about this: 1/2 + 1/4 + 1/8 + ..., you will get a constant
(1 vote)
• we know when we do definite integrals, we fine the area under a curve between two points. in that sense, what does this actually mean? how does summing up all the forces relate to finding the area of a function between two points .. i just cant relate these two things..
• good question

One way I find it useful to make sense of this in a simple way is to think of the units of the area as being equal to the units of the y axis (F) multiplied by the units of the x axis (r). Just as you would calculate area of any other surface.

So, now you are not just adding up all the forces; your integral is giving you y times x. In this case the amount of work done (Fr). (Think about it as each ,very thin column as having an area equal to F times delta r)

The area under a F/x curve is the work done by the force in moving from one point to the other. and is given by the intergal of the function F=f(x) between two points (or posiitons)

You can use the hookes law line F against extension to confirm this idea (energy stored in a spring = area under the curve = 1/2 Fx)
OK??
• if the test charge is at infinite height from the infinitely charged plate then would there be an electric field from the plate ??
• Yes, the strength of the field from a infinite plate with a uniform charge on its surface is the same regardless of the distance.
• lets say that there is a positive charged particle above the infinite plate. Will the particle move up infinitely because of the force produced by the constant electric field?