Work required to separate parallel plate capacitors

AI Thread Summary
The discussion centers on calculating the work required to separate parallel plate capacitors after tripling the plate separation. The user initially misapplied the formula for change in energy (ΔU) by reversing the initial and final energy terms, leading to an incorrect answer. It is emphasized that the new capacitance must be used in all calculations after the separation change. Clarifications were provided on properly substituting values in the energy equations. The user expressed gratitude after understanding the corrections needed for their solution.
avolp
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Homework Statement


A parallel plate capacitor with capacitance C is charged to a value q and then isolated. The separation between the plates is then tripled. What was the work required to separate the capacitors?

Homework Equations


U=1/2CV^2
Q=CV
C=ε0A/d

The Attempt at a Solution


W=ΔU=1/2C1V1^2-1/2C2V2^2
=1/2C*(Q/C)^2-1/2*(C/3)*(Q/C)^2
=Q^2/3C
I tried this but this wasn't the correct answer and I can't figure out why. Thank you in advance for the help
 
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avolp said:

Homework Statement


A parallel plate capacitor with capacitance C is charged to a value q and then isolated. The separation between the plates is then tripled. What was the work required to separate the capacitors?

Homework Equations


U=1/2CV^2
Q=CV
C=ε0A/d

The Attempt at a Solution


W=ΔU=1/2C1V1^2-1/2C2V2^2
=1/2C*(Q/C)^2-1/2*(C/3)*(Q/C)^2
=Q^2/3C
I tried this but this wasn't the correct answer and I can't figure out why. Thank you in advance for the help

Hi avolp, welcome to Physics Forums.

When you write your formula for ΔU you want to take the final energy and subtract the initial energy. You've got them the other way around, so your answer will be the negative of the correct value.

When you write the energy for the capacitor with the plate separation increased by a factor of 3, be sure to substitute the new capacitance for all the instances of the capacitor value C.
 
Last edited:
Oh okay. Now I understand. Thank you so much.
 
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