A parallel plate capacitor with an air gap between two plates is connected to a

Question

A parallel plate capacitor with an air gap between two plates is connected to a 6V battery. After charging, the energy stored in the capacitor is 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. Each plate has an area of 50 cm^2. What is the charge on the positive plate of the capacitor after the dielectric is inserted? What is the magnitude of the electric field between the plates before the dielectric is inserted? What is the dielectric constant of the material?

Explanation / Answer

before dielectric

Q = C*V

U1 = 0.5*C*V^2 = 0.5*Q*V = 0.5*Q^2/C

after the dielectric is placed

C' = kC

Q' = C'*V = k*Q

U' = k*U

U = 0.5*C*V^2

72*10^-9 = 0.5*C*6^2

C = 4*10^-9 F

k = U'/U = 317/72 = 4.402

(a)

Q' = C'*V = k*C*V

Q' = 4.402*4*10^-9*6

Q' = 1.056*10^-7 C

---------

(b)

E = Q/A*eo = (4*10^-9*6)/(50*10^-4*8.85*10^-12)

E = 54237.88 N/C

(c)k = U'/U = 4.402

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