How Can Capacitance Be Halved in a Parallel-Plate Capacitor?

AI Thread Summary
To halve the capacitance of a parallel-plate capacitor, one can increase the plate separation or decrease the plate area. The capacitance is directly proportional to charge, meaning reducing the charge does not affect capacitance in the same way. Therefore, halving the charge does not contribute to reducing capacitance. The discussion confirms that capacitance decreases with increased separation and decreased area, aligning with the formula C = Q/V. Understanding these relationships is crucial for manipulating capacitor behavior in practical applications.
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Homework Statement



Consider a charged parallel-plate capacitor. How can its capacitance be halved?
Check all that apply.

Double the charge.
Double the plate area.
*Double the plate separation.
*Halve the charge.
*Halve the plate area.
Halve the plate separation

Homework Equations



C = Q/V


The Attempt at a Solution



I know for sure that the capacitance can be reduced if the area decreases and separation increases. The part I am unsure is if the capacitance is decreased if the charge is decreased, although I am leaning more towards it does, since there will be less charge spread around the area of the plate. Thanks in advance. :smile:
 
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The part I am unsure is if the capacitance is decreased if the charge is decreased

The relevant equation you posted answers this
 
So capacitance is directly proportional to charge.
 

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