A parallel plate capacitor with an air gap is connected to a 6V battery. The cha

Question

A parallel plate capacitor with an air gap is connected to a 6V battery. The charged capacitor stores energy of 72nJ. Without disconnecting the capacitor from the battery, a dielectric material is inserted into the air gap and the energy stored in the capacitor is increased to 317nJ. A conducting plate in the capacitor has an area of 50 cm^2 (a) What is the charge on the positive plate of the capacitor after the dielectric is inserted? (b) What is the magnitude of the electric field between the plates before the dielectric is inserted? (c) What is the dielectric constant of the material?

Explanation / Answer

a)

final energy stores in the capacitor

Uf=72+317=389 nJ

since Uf=(1/2)QfV

(389*10-9)=(1/2)*Qf*6

Qf=129.67 nC

b)

Initial energy stored in capacitor

Ui=(1/2)CV2

since C=eoA/d

and V=Ed

U=(1/2)(eoA/d)(Ed)2=(1/2)*eo*A*E*V

E=2Ui/eoAV=2*(72*10-9)/(8.8542*10-12)*(50*10-4)*6

E=5.42*105N/C

c)

Uf/Ui =(1/2)CfV2/(1/2)CV2=KC/C =K

K=389/72=5.4

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