A parallel plate capacitor with air in the gap between the plates is connected t

Question

A parallel plate capacitor with air in the gap between the plates is connected to a 6 volt- battery. After charging, the energy stored in the capacitor is 64 nJ. Without disonnecting the capacitor from the battery, a dielectric is inserted into the gap and an additional 400 nJ of energy flows from the battery tot he capacitor.

A. What is the dieletric constant of the dielectric? Answer is 7.25

B. If each of the plates has an area of 50cm^2, what is the charge on the positive plate of the capacitor after the dielectric has been inserted? Answer is .155microC

C. What is the magnitude of the electric field between the plates before the dielectric is inserted? Answer is 4.82*10^5 N/C

D. What is the magnitude of the electric field between the plates after the dielectric is inserted? Answer is 4.82*10^5 N/C

Explanation / Answer

a) E = 1/2 C V^2

so when dielectric is put in C = k C0

so initially

64.0E-9 = 0.5*C*6^2

C0=3.56E-9 F

after its put in

464.0E-9 = 0.5*C*6^2

c = 2.58E-8 F

so k = C/C0 = 2.58E-8/3.56e-9= 7.25

B) Q = C V = 2.58E-8*6 = 1.55E-7 C = 0.155 microC

c) E = V/d

C = e0 A/d

so
3.56E-9 = 8.85E-12*50.0E-4/d
so d = 1.24E-5 m

E = 6/1.24E-5= 4.82E5 N/C

d) d doesnt change so E doesnt

E = 4.82E5 N/C

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