An air-filled parallel-plate capacitor has an area of1.4 m 2 and a capacitance o

Question

An air-filled parallel-plate capacitor has an area of1.4 m2 and a capacitance of 2.2 nF.

1. What is the distance between the plates? (mm)

2. If the capacitor is connected to a 32 V battery, what is the electric field inside the plates?

3. If the capacitor is connected to a 32 V battery, what is the magnitude of the charge on each plate?

4. When the capacitor is connected to a 32 V battery, what is the energy stored in the capacitor?

5. The battery is disconnected from the capacitor so that the charge stays on each plate. You push the plates inwards so that the distance between the plates is half of what it was. What is the energy stored on the plates now?

6. How much work (positive or negative) are you doing on the plates moving them inwards? Why does this work make sense in terms of the force and displacement you apply?

Explanation / Answer

1) we know that

C = A*epsilon/d

d = A*epsilon / C = 1.4*8.854e-12 / 2.2e-9 = 5.63e-3 m = 5.63mm

2) we know that

V = E*dx

E = V/dx = 32/5.63e-3 = 5.68e3 N/C

3) q = CV = 2.2e-9 * 32 = 70.4e-9 C = 70.4nC

4) U = 1/2 CV^2 = 1/2 * 2.2e-9 * 32^2 = 1126.4e-9 J

5) new capacitance = C' = A*epsilon / (d/2) = 2* A*epsilon/d = 2C

U = 1/2 * q^2/C' = 1/2 * q^2/2C = 1/2 * (70.4e-9)^2 / 2*2.4e-9 = 563.2e-9J

6) dU = 563.2e-9 - 1126.4e-9 = -563.2e-9J is the net change in energy

so work done on system = change in energy = 563.2e-9 J

WD = F*s = qE * s = (70.4e-9)/2 * 5.69e3 * (5.63e-3)/2 = 563.8e-9 J which is equal to the net energy change

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