Charged capacitor, electric field

  • #1

Homework Statement


A capacitor consists of two large metal disks of radius 2.4 meters placed parallel to each other, a distance of 0.7 millimeters apart. The capacitor is charged up to have an increasing amount of charge +Q on one disk and -Q on the other. At about what value of Q does a spark appear between the disks?


Homework Equations


E = [Q/(A/2(epsilon))](1- Z / (R^2+Z^2)^1/2)
R>>Z

E = [Q / (pi(r^2) / 2 * 8.85E-12)]

The Attempt at a Solution


I know that since one disk is positive and one disk is negative, the field points in one direction, so it should be E1 + E2. I tried plugging it into the equation, but then I'd still have 2 variables, E and Q. How do you find the value of Q when a spark appears? Is there a value for this that they did not provide?
 
Last edited:

Answers and Replies

  • #2
collinsmark
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look up "dielectric breakdown." I'm assuming there is an air gap between the plates (as opposed to a vacuum).
 
  • #3
Thanks for that. It says that sparks occur when the electric field strength is 3x10^6 N/C.

So if you have that, then:

E = E1 + E2

E = [Q/(A/2e)] + [Q/(A/2e)] (R>>Z)

3x10^6 = 2[Q/(A/2e)]

3x10^6 = 2[Q/(pi(2.4)^2/2(8.58E-12)]

Q = 3E20 N/C.

What am I doing wrong here?
 
  • #4
collinsmark
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It's difficult for me to interpret your math the way it is written. Perhaps if you could more carefully writing it out, including the units (when appropriate), it would help. But I think you might be dividing by something when you should be multiplying, or vise versa.

One other minor thing I noticed is that the constant you are using for the permittivity of free space is a little off.
 
  • #6
collinsmark
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Okay, you have approximated the electric field of a single plate as
[tex] E = \frac{Q}{\frac{A}{2\epsilon _0}} [/tex], and that's where i think you went wrong.

Try using Gauss' law again to model the electric field caused by a large plate, or find the correct equation in your textbook.
 

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