Question #166128

A parallel plate capacitor has an area of 5.00 cm 2 and a capacitance of 3.50 pF. The

capacitor is connected to a 12.0 V battery. After the capacitor is completely charged up, the

battery is removed

a. What must be the separation of the plates and how much energy is stored in the electric

field between the plates?

b. If a sheet of Mica with εr=3.5 is placed between the plates, how will the voltage across

the plates and the charge density on the plates change?

c. If the sheet of Mica has an area of 3.00 cm 2 so that it does not cover the whole area of

the plates, what would be the effective capacitance of this arrangement?

d. If the Mica sheet is removed and the plates are moved 1.00 mm farther from each other,

what will be the voltage between the plates, and how much work will it take to move the

plates apart?

Expert's answer

a. Find separation and energy:

b. The new charge and capacitance will be 3.5 times greater, so

The voltage:

c. The effective capacitance of this arrangement is the same as that of two capacitors in parallel, one with air and one with mica:

d. The capacitor is not connected to the voltage source anymore, so, the charge will not change. The new capacitance:

Voltage:

Work:

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