Work Done on a Parallel Plate Capacitor

AI Thread Summary
An isolated 5.60 microfarad parallel-plate capacitor with a charge of 4.90 mC has its capacitance reduced to 2.00 microfarads by changing the distance between the electrodes. The initial voltage is calculated to be 875 V using the formula C = Q/V. The main confusion arises regarding how to determine the distance between the plates in the final condition and which values remain constant during the separation. It's suggested to use the relationship V = Ed, where E is the electric field, to analyze changes as the plates are separated. Understanding how the electric field behaves during this process is crucial for solving the problem.
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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.

Homework Equations


W = Fd
c = Q/V

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:
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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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