All wrong
Who says the capacitor is of the commercially available type so well known to all of us?
Let's step back and look at the
equation for a plate capacitor:
C = (epsilon*A) /d where A is the area of one plate and d is the distance, epsilon being the permittivity of the medium (e.g. air).
We also have the equation C = Q/V where Q is the charge on the plates (on one plate as the othe rplate has the same charge with inverse sign) and V is the voltage.
We can combine both equations to Q/V = (epsilon*A)/d
Now let's assume we have a primitive capacitor made from two conductive (metal) plates at a distance d=d1 (whatever the value of d1 is doesn't matter). Let's further assume the size of the plates is constant (A=const), as well as the permittivity of the medium between the plates (epsilon = const). As there is no power source, there will be noc change in cahrge, therefore Q=const, too.
We are left with an equation with two variables: V and d. We know that V shall change from 9 V to 24 V, that leaves us with a corresponding change in d.
In plain Englisch: when the distance between the plates of a capacitor is changed, the voltae changes, too, all other parameters being the same.
I'll leave the plain math to the op.
Of course, this answer is only particularly correct, as are all other proposed solutions, as one will need a kind of power source to change the distance between the plates (or to walk across a floor or to...).
The implicit assumption here is that by "no power source" no "
electrical power source was meant.
