Parallel plate capacitor

  • Thread starter ledwardz
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  • #1
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Homework Statement



A parallel plate capacitor has a charge of 50pC
area = 10mm^2
dielectric material permittivity = 2.5
distance between plates = 0.5 mm

find the electric field
the potential difference between the 2 plates
the energy stored in the capacitor


Homework Equations



electric field = Charge density / permittivity
charge density = q / a

therefore E = 5*10^-3 / 2.5 = 2* 10 ^-3

PD= Ed = 2*10^-3 * 0.005 = 1*10^-5

energy u = 0.5V*q = 0.5 * 1*10^-5* 50*10^-9 = 2.5*10^-13 J

can anyone tell me if this is correct? thanks again for any help. Cheers, Lee:shy:
 

Answers and Replies

  • #2
gneill
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You seem to have lost track of the orders of magnitude in the charge density. Check the calculation.

The permittivity κ is a multiplier for the vacuum permittivity, εo.
 
  • #3
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You seem to have lost track of the orders of magnitude in the charge density. Check the calculation.

The permittivity κ is a multiplier for the vacuum permittivity, εo.

ahh pooo. so ε = εor but i have the right idea for the rest of the equation?

thanks for reply also.
 
  • #4
gneill
Mentor
20,925
2,867
ahh pooo. so ε = εor but i have the right idea for the rest of the equation?

thanks for reply also.

Sure. Straighten out the powers of ten and the permittivity and you're golden.
 

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