Circuit with a parallel plate capacitor

[strawberrysk8]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

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?

 

[LowlyPion]

The capacitor is disconnected from the battery and the plates are returned to their initial spacing.

Where does charge come from after the battery is disconnected?

 

[strawberrysk8]

i guess there is no new charge so is the charge still 16.2nC?

 

[strawberrysk8]

if it is...why is V = Q/C = 11.2V? If C = AE/d = 2.01(8.85*10^-12)/0.0124
According to the answer key it is supposed to be 2.23V using the formula initial VC = final VC.

 

[LowlyPion]

i guess there is no new charge so is the charge still 16.2nC?

Yes. That's what it looks like to me.

 

[LowlyPion]

if it is...why is V = Q/C = 11.2V? If C = AE/d = 2.01(8.85*10^-12)/0.0124
According to the answer key it is supposed to be 2.23V using the formula initial VC = final VC.

Wasn't the question just how much charge?

The voltage will drop when the area increases if the charge is the same.

 

[strawberrysk8]

oh okay. thank you so much!