A parallel-plate capacitor has capacitance C = 12.5 pF when the volume between t

Question

A parallel-plate capacitor has capacitanceC = 12.5 pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00 cm. The capacitor is connected to a battery and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted the charge on each plate has magnitude 45.0 pC

1. What is the dielectric constant K of the dielectric?

2.What is the potential difference between the plates before the dielectric has been inserted?

3.What is the potential difference between the plates after the dielectric has been inserted?

4.What is the electric field at a point midway between the plates before the dielectric has been inserted?

5.What is the electric field at a point midway between the plates after the dielectric has been inserted?
pC

Explanation / Answer

a)

Charge on the capacitor before dielectric insertion

Qbefore=CV

Charge on the capacitor after dielectric is inserted

Qafter=KCV

=>K=Qafter/Qbefore=45/25

K=1.8

b)

Potential difference between the plates before the dielectric is inserted is

V=Qbefore/Cbefore =25/12.5

V=2 Volts

c)

Since battery is not disconnected ,potential difference between after dielectric is inserted remains same ,so

V=2 Volts

d)

Area

A=pi*r2=pi*(0.03)2=2.83*10-3 m2

The electric field is same at all points between the plates .so electric field at a point midway between the plates before the dielectric has been inserted

Ebefore=Qbefore/Aeo=(25*10-12)/(2.83*10-3)(8.85*10-12)

Ebefore=999.1 N/C

e)

electric field at a point midway between the plates after the dielectric has been inserted is

Eafter=Qafter/Aeo =(45*10-12)/(2.83*10-3)(8.85*10-12)

Eafter=1798.4 N/C

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