Increase in Charge on a Parallel Plate Capacitor

  • Thread starter Drakkith
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  • #1
Drakkith
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Homework Statement


The charge on the 3.00 cm2 area plates of an air-filled parallel plate capacitor is such that the electric field is at the breakdown value. By what factor will the maximum charge on the plates increase when bakelite is inserted between the plates? (The dielectric strength of air is 3.00 ✕ 106 V/m and that of bakelite is 2.40 ✕ 107 V/m. The dielectric constant of bakelite is 4.90.)

Homework Equations


C=Q/V
C=kC0
C00A/d
V=Ed

The Attempt at a Solution


I'm not sure what to do here. I don't even have a strategy to tackle this problem. So far I've tried manipulating all the equations I can find that have to do with capacitors, but it hasn't gotten me anywhere.
 

Answers and Replies

  • #2
Drakkith
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Okay, I managed to solve it.

If Q=CV, then Q0=C0V0 and Q1 = C1V1.
Since V0 is the breakdown voltage of air, Q0 = 3x106C0.
C1=4.9C0, so Q1 = 4.9C0V1. But V1 is the dielectric breakdown voltage.
So Q1=(4.9)(2.4x107)C0.
The ratio of increase is Q1/Q0, so that becomes 1.176x108C0/3x106C0.
C0 cancels out of the top and bottom and we get Q1/Q0 = 39.2.
 
  • #3
gneill
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That works. You could also have approached it via the electric field strength given the charge density on the plates, ##E = \frac{\sigma}{\epsilon_o}##, as the charge is proportional to the charge density and the plate area is fixed. Rearrange as ##\sigma = \epsilon_o E## and remember that the dielectric constant multiplies ##\epsilon_o## for the bakelite. Form a ratio for the two materials and you're done.
 
  • #4
Drakkith
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Thanks gneill!
 

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