Work required to separate parallel plate capacitors

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SUMMARY

The work required to separate parallel plate capacitors is calculated using the energy difference formula ΔU = U_final - U_initial. The initial energy is given by U_initial = 1/2 C V^2, while the final energy after tripling the plate separation is U_final = 1/2 (C/3) V^2. The correct expression for work done is W = Q^2 / (3C), but the initial and final energies must be correctly identified to avoid negative results.

PREREQUISITES
  • Understanding of capacitor fundamentals, including capacitance and energy storage.
  • Familiarity with the equations U = 1/2 CV^2 and Q = CV.
  • Knowledge of how capacitance changes with plate separation, specifically C = ε0A/d.
  • Basic algebra skills for manipulating equations and solving for variables.
NEXT STEPS
  • Review the derivation of energy stored in capacitors, focusing on U = 1/2 CV^2.
  • Study the effects of changing plate separation on capacitance and energy in parallel plate capacitors.
  • Explore the concept of electric potential energy in capacitors and its implications in circuit design.
  • Investigate practical applications of capacitors in electronic circuits, particularly in energy storage and timing applications.
USEFUL FOR

Students studying electromagnetism, electrical engineers, and anyone involved in designing or analyzing circuits with capacitors.

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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