Please show all work. Thank you. The plates of a parallel-plate capacitor in vac

Question

Please show all work. Thank you. The plates of a parallel-plate capacitor in vacuum are 5 mm apart and 2 m2 in area each. The capacitor is connected to a 120 V battery. Compute . V a) the energy stored in the capacitor t vb) While the capacitor remains connected to the battery, a dielectric of dielectric constant K 3 is inserted between the plates of the capacitor. Describe how (increase or decrease and by what factor) the 1) capacitance, 2) voltage, 3) charge on each plate, and 4) electric energy stored will change. c) How would the above quantities change if instead the dielectric is inserted after the capacitor gets charged and then disconnected from the battery.

Explanation / Answer

a)

Capacitance

C=eoA/d =(8.8542*10-12)*(2)/(5*10-3)=3.54*10-9F

Energy stored in capacitor is

E=(1/2)CV2=(1/2)*(3.54*10-9)*1202=2.55*10-5J

b)

1.

Capacitance is

C' =KC=3*3.54*10-9=1.062*10-8F

So Capacitance increases by a factor of 3

2.

Voltage stays the same ,since it is connected to battery

3.Charge increases by a factor of 3 ,since Q=CV

4.

Energy stored increases by a factor of 3 ,sine

E=(1/2)CV2

c)

1.

Capacitance is

C' =KC=3*3.54*10-9=1.062*10-8F

So Capacitance increases by a factor of 3

2.

Voltage decrease by a factor of 3 ,since

V=Q/C

3.Charge stays the same ,since it fully charged

4.

Energy stored decreases by a factor of 3 ,sine

E=(1/2)(Q2/C)

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