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Parallel plate capacitor electric charge

  • #1
Consider an air filled parallel plate capacitor with plate area A and gap width d. The plate charge is Q. The total energy stored in the capacitor is given by U= Q^2 *d /2 E_o A. With the battery connected, fill the gap by a slab with the dielectric constant k. Given : E= 70 V, k= 3.8, d=0.8 mm and A = 25.6 cm^2, E_o = 8.85 x 10^-12 C^2/Nm^2, find the electric charge on the plate. Answer in units of C.
I really have no idea how to do this. I know that I have everything needed to find the total energy, but I don't really know how to get the charge from that. Can someone help?
 

Answers and Replies

  • #2
berkeman
Mentor
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What is the equation relating Q, C and V? What is the equation relating C, epsilon, A and d?
 
  • #3
I figured it out... thank you.
 
  • #4
1
0
horribly similar is: given only area of each plate, a distance between the plates, and instead, k = 1 (due to a vacuum as opposed to a slab), and the voltage, how would one find surface charge density (in Coulombs/meter^2)? I think i'm only confusing myself by changing various units around (such as volts to Joules/Coulomb)

So the work i've done so far includes:
I've noticed this should be relatively easy; the formula for density is simply just magnitude of the electric field multiplied with the surface area.
First I converted surface area from cm^2 to meters^2, I'm simply stuck on changing the charge difference from volts to coulombs :(
 
Last edited:

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