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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.