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Work Done on a Parallel Plate Capacitor

  • Thread starter vertabatt
  • Start date
30
0
1. Homework Statement
An isolated 5.60 microfarad parallel-plate capacitor has 4.90 mC of charge. An external force changes the distance between the electrodes until the capacitance is 2.00 microfarads.


2. Homework Equations
W = Fd
c = Q/V


3. The Attempt at a Solution
My first instinct was to solve to the potential of the system in it's initial conditions:

C = Q/V

5.60*10^-6 F = .0049C/V
V = 875 V

Now is where I become a little confused. I'm not sure how I can solve for the distance between the plates in the final condition. Also, I'm not sure what values stay constant as the plates are separated.

Any help is much appreciated!
 
Last edited:

Answers and Replies

Doc Al
Mentor
44,827
1,083
Use C = Q/V. Note that V = Ed, where E is the field between the plates. Does the field change as the plates are separated?
 

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