Part A: A capacitor is completely charged with 650 nC by a voltage sourcethat ha

Question

Part A:
A capacitor is completely charged with 650 nC by a voltage sourcethat 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 withthe voltage 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 is thestored energy if the air gap is now 3 mm?

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

C= (0A)/(d1)
or
A=(Cd)/(0) = ((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'= (0A)/(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 J NotRight






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

Need some help!

Explanation / Answer

Part A The capacitance is Q = C * V where C = 650 nC = 650 * 10-9 C and V = 275 V Part B The capacitance of the capacitor is given by C = (oA/d) where o is the permitivity of free space,A isthe area of the plates and d is the distance between theplates. Part C 1)The energy stored in the capacitor is u = (1/2)oE2 where E = (V/d),the initial and final distance between theplates of the capacitor is 7 mm and 3 mm respectively. 2)The capacitance when d = 3 mm = 3 * 10-3 mis C' = (oA/d') 3)The potential energy with this air gap is U = (1/2)C' * V2 where E = (V/d),the initial and final distance between theplates of the capacitor is 7 mm and 3 mm respectively. 2)The capacitance when d = 3 mm = 3 * 10-3 mis C' = (oA/d') 3)The potential energy with this air gap is U = (1/2)C' * V2

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.