Work Done on a Parallel Plate Capacitor

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SUMMARY

The discussion centers on the calculations involving an isolated 5.60 microfarad parallel-plate capacitor with a charge of 4.90 mC, which is altered to a capacitance of 2.00 microfarads. The initial voltage across the capacitor is calculated to be 875 V using the formula C = Q/V. The main confusion arises regarding the determination of the distance between the plates in the final condition and the constancy of values as the plates are separated. The relationship between voltage, electric field, and distance is highlighted, specifically V = Ed.

PREREQUISITES
  • Understanding of capacitor fundamentals, specifically parallel-plate capacitors.
  • Familiarity with the equations for capacitance (C = Q/V) and electric field (E = V/d).
  • Knowledge of the relationship between charge, voltage, and capacitance.
  • Basic algebra skills for manipulating equations to solve for unknowns.
NEXT STEPS
  • Explore the concept of electric field strength in capacitors and how it varies with distance.
  • Study the effects of changing capacitance on voltage and charge in parallel-plate capacitors.
  • Learn about energy stored in capacitors and the work done in changing configurations.
  • Investigate the implications of separating capacitor plates on capacitance and electric field.
USEFUL FOR

This discussion is beneficial for physics students, electrical engineering students, and anyone studying capacitor behavior in circuits and electrostatics.

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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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