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

[Heat]

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.

 

[blochwave]

The part I am unsure is if the capacitance is decreased if the charge is decreased

The relevant equation you posted answers this

 

[Heat]

So capacitance is directly proportional to charge.