Two square plate electrodes are each 10.0 cm long on each side. They are placed

Question

Two square plate electrodes are each 10.0 cm long on each side. They are placed so that they are separated by 0.100 mm so that they form a parallel plate capacitor.
a) What is the capacitance of this capacitor?
b) If the plates are connected by copper wires to the terminals of a 24 V battery what is the voltage difference between the plates? What is the charge on each plate?
c) Now, using insulating handles (but leaving them connected to the battery) the plates are pulled apart to a distance of 1.00 mm. What is the new voltage difference between the plates? What is the new charge on the plates?
d) The plates are returned to their original separation. They are disconnected from the battery but not discharged. Using the insulating handles they are again pulled apart to a distance of 1.00 mm. Now what is the voltage difference between the plates? What is the new charge on the plates?

Explanation / Answer

Part A)

The formula to determine the capacitance of parallel plates is

C = oA/d

C = (8.85 X 10-12)(.1)(.1)/(1 X 10-4)

C = 8.85 X 10-10 F   which is 885 pF

Part B)

Since the plates are connected to a 24 volt battery, the voltage difference betowwn the plates is 24 V

To find Q, we use Q = CV

Q = (8.85 X 10-10)(24) = 2.124 X 10-8 C which is 21.24 nC

Part C)

When the plates are pulled apart, they will have a new capacitance

C = oA/d

C = (8.85 X 10-12)(.1)(.1)/(1 X 10-3)

C = 8.85 X 10-11 F   which is 88.5 pF

The voltage stays at 24V since it is still attached to the battery

The new Q is

Q = CV = (8.85 X 10-11)(24) = 2.124 X 10-9 C which is 2.124 nC

Part D)

When the plates are disconneted from the battery, the total amount of charge can not change. No supply to make the change, so the Q remains at the original value of 2.124 X 10-8 C  or 21.24 nC.

Now they are again pulled apart, so the Capacitance is 8.85 X 10-11 F

Then using Q = CV, we can find the new V

V = Q/C = (2.124 X 10-8)/(8.85 X 10-11) = 240 V

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