Parallel plate capacitors

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Homework Statement



A parallel-plate capacitor has plates with an area of 445 cm2 and an air-filled gap between the plates that is 1.51 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery. There is 4.311536113 x 10 ^-5 J of energy stored in the capacitor. The separation between the plates is now increased to 3.02 mm. How much energy is stored in the capacitor now?

Homework Equations



I know that since the capacitor is disconnected from the battery, V can change with C while Q, being trapped in the disconnected
capacitor, is constant. I calculated Q for the original
capacitor, which was about 1.5 x 10^-7 C. When the spacing, d, between the plates is increased, C changes, and V changes with it, so I know that I can't use the equation U=(1/2)CV^2 (with V=575 V) to find the energy, but I'm not sure exactly what I can use.
 

Answers and Replies

  • #2
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...When the spacing, d, between the plates is increased, C changes, and V changes with it...
So let's see how they change.
 
  • #3
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Well, I thought that maybe by treating the increased space as a dielectric field and using the dielectric constant I could figure out the new values of V and C and use the Q to calculate the new Energy, but apparently I wasn't on the right track...
 
  • #4
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How is the capacitance related to the distance between the plates?
 
  • #5
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In calculating the capitance, the equation is C=(constant) x area/distance. So, b/c the battery is disconnected and the plates are pulled further away from each other, there needs to be a new equation for capitance...but I'm not sure what
 
  • #6
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...there needs to be a new equation for capitance...but I'm not sure what
Not a new equation. Only a new value for one parameter - the distance between the plates.

By how many times does the distance between the plates increase? How will this affect the capacitance? Will it become greater or smaller? By how many times?
 

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