Parallel plate capacitor with changing spacing, voltage, and dielectric

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strawberrysk8
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1.) Consider a parallel plate capacitor with plates of area 1.36 m2 whose plates are 1.24 cm apart. The gap between the plates is filled with air (assume that κair is unity) and the positive plate has a charge of 10.5 nC on it while the negative plate has a negative charge of equal magnitude on it.

The capacitor has its plate-spacing reduced to 1.12 mm, and the plates are connected to a 1.51 V battery. So that the charge on the positive side of the capacitor is now 16.2nC.

The capacitor is disconnected from the battery and the plates are returned to their initial spacing. So that the voltage across the capacitor is now 16.7V.

The space between the plates is now filled with a fluid with a dielectric constant of 5.07.
So that the voltage across the capacitor is now 3.297V.

The area of overlap between the capacitor plates is increased to 2.01 m2.

What is the charge on the positive plate of the capacitor?
 

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Please show your work and where you get stuck so that we can help you.
 


V = Q/C
C = AE/d

Q = CV = AEV/d
d = 0.00112m
V = 1.51V
A = 1.36
Q = 16.2nC

V = Q/C = Q/ (AE/d) = Qd/AE
d = 0.0124m
Q = 16.2nC
A = 1.36m2
V = 16.7V

V = V(0)/k
V(0) = 16.7V
k = 5.07
V = 3.297V

Q = CV = AEV/d
A = 2.01m2
V = 3.297V
d = 0.0124m
Q = 4.73nC
BUT Q IS WRONG? WHY IS IT WRONG?
 


Trick question. The charge on the plates can't change without an external source to drive it. The charge stays the same when the area increases (although the charge density decreases since the area increases).