Physics Help Forum Capacitors and electric fields

 Feb 21st 2010, 09:56 PM #1 Junior Member   Join Date: Feb 2010 Posts: 9 Capacitors and electric fields 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? Attempt: When they mean sparks, they mean "dielectric breakdown", right? I looked it up, and it says that that occurs around E = 3E6 N/C. E = [Q/(A/2ε)](1-(Z/sqrt(R^2+Z^2)) Because Z<
Feb 22nd 2010, 02:59 PM   #2
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 Originally Posted by ZergLurker 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? Attempt: When they mean sparks, they mean "dielectric breakdown", right? I looked it up, and it says that that occurs around E = 3E6 N/C. E = [Q/(A/2ε)](1-(Z/sqrt(R^2+Z^2)) Because Z<
The formula used for capacitance in the above equations is not correct since it does not include plate separation distance (d). And breakdown voltage is not properly accounted for. Use the following approach instead...

Assuming that air at standard pressure is being used as the dielectric, the approximate breakdown voltage is about 30 kV/cm, or about 3 kV/mm. First, figure out what the breakdown voltage would be at a distance of 0.7 mm. That will be the appropriate voltage (V) across the capacitor.

You also need to calculate the capacitance so that you can calculate the amount of stored charge at the breakdown voltage V above. Since the dielectric is air, we know that the relative dielectric constant is 1, so the formula for capacitance simplifies to:

C = εA/d (in SI units), where ε = 8.854×10−12 F m–1

Compute C using the area (A, in square meters) and plate separation (d, in meters). Then solve for Q as follows below. It should be much less than a coulomb.

Q = CV

This should do the trick for you!

Bert

Feb 24th 2010, 12:55 AM   #3
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 Originally Posted by BertHickman The formula used for capacitance in the above equations is not correct since it does not include plate separation distance (d). And breakdown voltage is not properly accounted for. Use the following approach instead... Assuming that air at standard pressure is being used as the dielectric, the approximate breakdown voltage is about 30 kV/cm, or about 3 kV/mm. First, figure out what the breakdown voltage would be at a distance of 0.7 mm. That will be the appropriate voltage (V) across the capacitor. You also need to calculate the capacitance so that you can calculate the amount of stored charge at the breakdown voltage V above. Since the dielectric is air, we know that the relative dielectric constant is 1, so the formula for capacitance simplifies to: C = εA/d (in SI units), where ε = 8.854×10−12 F m–1 Compute C using the area (A, in square meters) and plate separation (d, in meters). Then solve for Q as follows below. It should be much less than a coulomb. Q = CV This should do the trick for you! Bert
So

C = (8.854E-12 F/M)*(2.4^2*pi M^2) / (0.0007 M) = 1.602E-10 / 0.0007 M = 2.29E-7 F = 2.29E-7 C/V

Q = CV = 2.29E-7 (C/V) * 3E6 (V/M) = 6.87E-1 C/M

My units don't come out correctly. What exactly am I doing wrong?

Wait. Do I do: 3E6 V / M * 0.0007 M = 2100

Q = CV = 2.29E-7 * 2100 = 4.8E-4 C?

: It is! Thank you very much, good sir!

Last edited by ZergLurker; Feb 24th 2010 at 11:40 PM.

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