Charged capacitor, electric field

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SUMMARY

The discussion focuses on calculating the charge (Q) required for a spark to appear between two parallel metal disks of a capacitor, with a radius of 2.4 meters and a separation of 0.7 millimeters. The critical electric field strength for spark formation is established at 3x10^6 N/C. The participants suggest using the equation E = [Q/(A/2ε)] and emphasize the importance of correctly applying Gauss' law to model the electric field. The correct value of the permittivity of free space is also highlighted as crucial for accurate calculations.

PREREQUISITES
  • Understanding of capacitor physics and electric fields
  • Familiarity with Gauss' law
  • Knowledge of the permittivity of free space (ε₀)
  • Ability to manipulate equations involving electric field strength
NEXT STEPS
  • Review Gauss' law applications in electrostatics
  • Study the concept of dielectric breakdown in air
  • Learn about the calculation of electric fields for parallel plate capacitors
  • Examine the correct values and units for physical constants, including ε₀
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Physics students, electrical engineers, and anyone interested in understanding capacitor behavior and electric field calculations.

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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?
 
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look up "dielectric breakdown." I'm assuming there is an air gap between the plates (as opposed to a vacuum).
 
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?
 
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.
 
Okay, you have approximated the electric field of a single plate as
E = \frac{Q}{\frac{A}{2\epsilon _0}}, 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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