Maximum Working Voltage of a Capacitor

  • Thread starter tomrja
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  • #1
tomrja
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Homework Statement



A parallel-plate capacitor has plates with 38.1 cm2 area separated by a 29.0 μm layer of Pyrex glass.

(a) Find its capacitance.
(b) Find its maximum working voltage.

Kappa for pyrex glass is = 5.6
A= 38.1 cm^2 = 3.81*10^-3 m^2
d= 29*10^-6 m

Homework Equations



C=(Kappa*Epsilon*A)/d

Q=CV

The Attempt at a Solution



I have gotten the answer to part a by just plugging and chugging with the equation C=(Kappa*Epsilon*A)/d. For the second part I don't understand how to calculate the maximum voltage if I don't have a charge (Q). I'm sure there is some way to manipulate these equations. Any help?
 

Answers and Replies

  • #2
gneill
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I have gotten the answer to part a by just plugging and chugging with the equation C=(Kappa*Epsilon*A)/d. For the second part I don't understand how to calculate the maximum voltage if I don't have a charge (Q). I'm sure there is some way to manipulate these equations. Any help?

The maximum working voltage will be determined by the breakdown voltage of the dielectric. More specifically, the maximum field strength it can sustain before it breaks down and shorts the capacitor.

Dielectric strength is given in V/m, just like the electric field between the plates (hint, hint).

Look up the dielectric strength of Pyrex glass.
 
  • #3
tomrja
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Awesome! Thank you so much!
 

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