Question

Part A:
A capacitor is completely charged with 650 nC by a voltagesourcethat has 275 V.
What is its capacitance?

C=Q/V = (550e-9 C)/375V = 1/47e-9 F (1.47 nF)ConfirmedRight

Part B:
Now the plates of the charged capacitor are pushed together withthevoltage source already disconnected:

- The voltage drop between the plates decreases.
- The energy stored in the capacitor remains the same.
- The capacitance increases.
- The voltage drop between the plates increases.
- None of the above.


Part C:
The initial air gap of the capacitor above was 7 mm. What isthestored energy if the air gap is now 3 mm?

1) We know Capacitance when the air gap is 7 mm (d1=7e-3 m), butnotthe area, A. So use:

C= (ε0A)/(d1)
or
A=(Cd)/(ε0) = ((1.47E-9 F)*(7e-3 m))/(8.85e-12) = 1.163m^2


2) Using the area, A, that you find, find the capacitance, C'withan air gap of 3 mm. (d2=3e-3 m)

C'= (ε0A)/(d2) = ((8.85e-12)*(1.163 m^2))/(3e-3) = 3.431e-9F


3) The potential energy with this air gap of 3 mm is:

U = 0.5C'V^2 = 0.5(3.431e-9 F)(375^2) = 2.41e-4 JNotRight






Alright, as you can see, I worked out the last part completelyandI'm pretty sure my work is right. However, the answer is showingupas wrong. I'm not sure what I did wrong here. Also, I'm notsurewhat to do for Part B and I only have one try. Wouldn't theenergyremain the same (or nothing happen at all)?

Need some help!