3. A parallel plate capacitor C=10uF with distance between the plates 2um is att

Question

3. A parallel plate capacitor C=10uF with distance between the plates 2um is attached to a battery with voltage 12V. After the capacitor is fully charged, it is disconnected from the battery and a dielectric with K=2.4 is inserted between the plates of the capacitor, filling the whole space a. What is the area of the capacitor? Show work. b. What is the charge on the capacitor both before and after the dielectric is inserted. Explain. c. What is the electric field in the capacitor, before and after the dielectric is inserted. Explain d. What is the induced electric field in the dielectric due to polarization of the charges? e. What is the voltage across the capacitor, before and after the dielectric is inserted. Explain f. What is the energy stored in the capacitor, before and after the dielectric is inserted. Explain g. What work was done to insert the dielectric in the capacitor? Was the dielectric “shoved in" or was it "sucked in"? Explain

Explanation / Answer

(a) C = e0 A / d

10 x 10^-6 = (8.854 x 10^-12) (A) / (2 x 10^-6)

A = 2.26 m^2

(B) before = after charge = C V = 120 uC  

(C) before , E = V/d = 12 / (2 x 10^-6)

= 6 x 10^6 V / m

after , E' = E/ k = 2.5 x 10^6 V / m

(D) E_induced = E - E' = 3.5 x 10^6 V/ m

(E) before, V = 12 Volt

after, V = 3.5 x 10^6 x 2 x 10^-6 = 7 Volt

(F) before, Ui = Q V / 2

= 720 x 10^-6 J

after, Uf = Q V' / 2 = 420 x 10^-6 J

(G) Work done = Uf - Ui = - 300 x 10^-6 J

Sucked in.

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