If the magnitude of the electric field in air exceeds...

Name: if the magnitude of the electric field in air exceeds roughly 3 times 105 nc the air breaks down and a spark forms for a two disk capacitor of radius 52 cm with a gap of 3 mm what is the max 35808
Uploaded: 2020-06-19T18:26:32-08:00
Duration: 3 min 46 s
Channel: Jonathan Everett
Description: if the magnitude of the electric field in air exceeds roughly 3 times 105 nc the air breaks down and a spark forms for a two disk capacitor of radius 52 cm with a gap of 3 mm what is the max 35808

If the magnitude of the electric field in air exceeds roughly $3 \times 10^5$ N/C, the air breaks down and a spark forms. For a two-disk capacitor of radius 52 cm with a gap of 3 mm, what is the maximum charge (plus and minus) that can be placed on the disks without a spark forming (which would permit charge to flow from one disk to the other)? The constant $\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2)$.

Transcript

00:01 Hello, and welcome to digital tea with mr.

00:03 E, where we will see what's brewing in the world of physics solutions.

00:07 So in this particular problem, we learn that when you have an electric field of three times 10 to 6th, newton's per coulum, in a capacitor or two parallel plates that have air in the middle, you'll begin to get sparks.

00:24 So that's the point where air breaks down and electricity can start to spark through it.

00:29 At any rate, our task is to figure out how many electrons have to be moved from one plate to the other plate in order to create the electric field of 3 times 10 to 6 newtons per coolum and create sparks inside of our capacitor.

00:44 And we also know that the parallel plates are circular in nature with a diameter of 4 centimeters.

00:51 So the first thing i would like to do is to determine the area of the plates.

00:56 That's going to be pi times r squared.

01:00 And we have a diameter of four centimeters, so that would mean that we have a radius of two centimeters.

01:11 So we could go through and do pi times 0 .02 squared, and that would give us an area of 1 .26 times 10 to the minus 3rd meter squared.

01:26 Then from there, we can go with the standard derivation from gauss's law that says the perpendicular component of e times a is equal to the charge over epsilon sub -knot, where epsilon sub -not is a constant of 8 .85 times 10 to the minus 12.

01:49 Uh.....

Question

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