A capacitor is made of plates with an area of .15 m2 that are separated by 3 mm

Question

A capacitor is made of plates with an area of .15 m2 that are separated by 3 mm of mica (dielectric constant of 7.) The capacitor is charged by a 20 V battery, and then removed from the battery. The mica is then removed from between the plates, leaving air.

a. What was the capacitance before the mica was removed?

b. What is the capacitance after the mica was removed?

c. Consider just charge on the plates and the voltage across the plates. Which of them is the same before and after the mica is removed? Why?

d. Find the electric field between the plates after the mica is removed. [4.67x104 V/m]

e. Find the potential energy stored by the plates after the mica is removed. [4.34x10-6 J]

Explanation / Answer

a)

Capacitance is given by

Ck=KeoA/d =7*(8.8542*10-12)(0.15)/(3*10-3)

Ck=3.1*10-9 F or 3.1 nF

b)

Capacitance without dielectric

C =Ck/K =3.1*10-9/7 =4.427*10-10 F or 0.4427 nF

c)

Since battery is disconnected ,Charge remains same and Voltage across the plate changes

d)

Charge before mica was removed

Qbefore =CKVbefore=(3.1*10-9)(20)=6.2*10-8 C

since charge remains same

Vafter=Q/C =6.2*10-8/(4.427*10-10)=140 V

Electric field is

E=V/d =140/(3*10-3)

E=4.67*104 V/m

e)

Uafter=(1/2)CV2 =(1/2)(4.427*10-10)(140)2 =4.34*10-6 J

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