A parallel-plate air capacitor is made by using two plates 11 cm square , spaced

Question

A parallel-plate air capacitor is made by using two plates 11 cm square , spaced 3.2 mm apart. It is connected to a 12-V battery.

Part A What is the capacitance? Express your answer with the appropriate units. C = 33 pF Correct

Part B What is the charge on each plate? Express your answer with the appropriate units. Q = 400 pC Correct

Part C What is the electric field between the plates? Express your answer with the appropriate units. E = 3800 Vm Correct

Part D What is the energy stored in the capacitor? Express your answer with the appropriate units. U = 2.4 nJ Correct

Part E The battery is disconnected and then the plates are pulled apart to a separation of 7.4 mm. What is the capacitance in this case? Express your answer with the appropriate units. C = 14.5 pF correct

Part F What is the charge on each plate in this case? Express your answer with the appropriate units.

Part G What is the electric field between the plates in this case? Express your answer with the appropriate units.

Part H What is the energy stored in the capacitor? Express your answer with the appropriate units.

I have figured out part A to E, but not part F to H. Please help! Thank you!

Explanation / Answer

(A) C = e0 A /d

= (8.854 x 10^-12) (0.11^2) / (3.2 x 10^-3)

= 3.35 x 10^-11 F = 33.5 pF

(B) Q = C V = 33.5 pF x 12 V = 402 pC

(C) E = V / d = 12 / (3.2 x 10^-3)

= 3750 V/m

(d) U = C V^2 /2 =2.41 x 10^-9 J = 2.41 nJ

(e) C = (8.854 x 10^-12)(0.11^2)/(7.4 x 10^-3)

C = 14.5 x 10^-12 F = 14.5 pF

(f) battery is disconnected so charge will remain same on plates.

Q = 402 pC

(g) E = Q / e0 A

Q and A is same henceE will also remain same.

E = 3750 V/m

(h) U = Q^2 / 2C = 5.57 x 10^-9 J = 5.57 nJ

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