Parallel Plate Capacitor and Battery 1 2 3 4 5 6 7 Two parallel plates, each hav

Question

Parallel Plate Capacitor and Battery

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Two parallel plates, each having area A = 3223 cm2are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.44 cm.

1)

What is Q, the charge on the top plate?

C

2)

What is U, the energy stored in the this capacitor?

J

3)

The battery is now disconnected from the plates and the separation of the plates is doubled ( = 0.88 cm). What is the energy stored in this new capacitor?

J

4)

What is E, the magnitude of the electric field in the region between the plates?

N/C

Explanation / Answer

1) C = Aeo/d

= 3223*10^-4*8.85*10^-12 / 0.44*10^-2

= 6.48*10^-10 F

Q = CV = 6.48*10^-10*6 = 3.88*10^-9 C

2) E = 0.5CV^2

= 0.5*6.48*10^-10*6^2 = 1.16*10^-8 J

3) if the seperation of plates is doubled , Capacitance will be halved and volatge will be doubled

E = 0.5CV^2

C is halved and V is doubled

so E(new) = 2E

= 2*1.16*10^-8 = 2.33*10^-8 J

4) E = Q/Aeo

= 3.88*10^-9 / 3223*10^-4*8.85*10^-12

= 1360.28 N/C

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