Charged capacitor, electric field

In summary, the homework statement says that a capacitor consists of two metal disks that are placed parallel to each other and a distance of 0.7 millimeters apart. At about what value of Q does a spark appear between the disks? Gauss' law states that the electric field strength is proportional to the product of the magnitude of the electric field and the square of the distance between the two points.
  • #1
DeadFishFactory
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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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  • #2
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
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
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.
 

FAQ: Charged capacitor, electric field

What is a charged capacitor?

A charged capacitor is an electrical component that stores electrical energy in the form of an electric charge. It is made up of two conductive plates separated by a dielectric material, which can be charged by connecting it to a power source.

What is the electric field of a charged capacitor?

The electric field of a charged capacitor is the force per unit charge acting on a charge placed in the space between the plates. It is directly proportional to the amount of charge stored on the plates and inversely proportional to the distance between the plates.

How is the electric field of a charged capacitor calculated?

The electric field of a charged capacitor can be calculated using the equation E = Q/εA, where E is the electric field, Q is the charge stored on the plates, ε is the permittivity of the dielectric material, and A is the area of the plates.

What happens to the electric field of a charged capacitor when the distance between the plates is increased?

When the distance between the plates of a charged capacitor is increased, the electric field decreases. This is because the electric field is inversely proportional to the distance between the plates, so as the distance increases, the electric field weakens.

How does a charged capacitor affect an electric circuit?

A charged capacitor can affect an electric circuit in several ways. It can act as a temporary power source, supplying energy to the circuit when needed. It can also act as a filter, smoothing out fluctuations in the circuit's voltage. Additionally, a capacitor can store and release energy quickly, making it useful for certain types of electronic devices.

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