What Happens to Charge When Capacitor Plate Separation Doubles?

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SUMMARY

When the separation between the plates of a parallel plate capacitor is doubled while maintaining a constant potential difference, the charge on the plates is halved. This conclusion is derived from the relationship defined by the equations CV=Q and C=εoA/2d. The capacitance C is inversely proportional to the distance d, indicating that an increase in separation results in a decrease in charge. The initial confusion regarding the answer key was clarified, confirming that the charge does not double but rather is cut in half.

PREREQUISITES
  • Understanding of capacitor fundamentals, specifically parallel plate capacitors.
  • Familiarity with the relationship between capacitance, charge, and voltage (CV=Q).
  • Knowledge of the formula for capacitance: C=εoA/d.
  • Basic grasp of electric fields and potential difference in capacitors.
NEXT STEPS
  • Review the derivation of capacitance for parallel plate capacitors.
  • Explore the effects of varying potential difference on charge in capacitors.
  • Learn about energy stored in capacitors and its relation to charge and voltage.
  • Investigate real-world applications of capacitors in electronic circuits.
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Students studying physics, particularly those focusing on electromagnetism, as well as educators seeking to clarify concepts related to capacitors and charge behavior.

physgrl
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Homework Statement



If the separation between the plates of a parallel plate capacitor is doubled and the potential difference is held constant by a battery, the magnitude of the charge on the plates will:

*a. double

b. quadruple

c. be cut in half

d. not change

Homework Equations



CV=Q
C=εoA/2d

The Attempt at a Solution



CV=Q
C proportional to Q
εoA/2d proportional to Q
1/d proportional to Q

now if the separation d is doubled the charge Q should be cut in half instead of doubled (as the answer key says) right? so have I made a mistake? Is it supposed to be done differently?
 
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Hi physgrl! :smile:

You are right and the answer key is wrong.

Btw, I have a different formula for the capacitance of 2 parallel plates.
It's ##C={\epsilon A \over d}##, as you can see on the wiki page: http://en.wikipedia.org/wiki/Capacitance

But that does not change your line of reasoning, which is correct.
 
ohh yeah the 1/2 was a mistake...thanks
 

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