A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate

Question

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 6.00 mm and dielectric constant k = 2.00. The capacitor is connected to a battery that creates a constant voltage V = 5.00 V . Throughout the problem, use 0 = 8.85×1012 C2/Nm2

A: Find the energy U1 of the dielectric-filled capacitor. Express your answer numerically in joules

B: The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric. Express your answer numerically in joules.

C.The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.Express your answer numerically in joules.

D:In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric?Express your answer numerically in joules

Explanation / Answer

Let's calculate the capacitance of capacitor

C= K0A/ d ( where A = area, d = separation)

C= (2)( 8.85×10^12)( 10 x 10^-4)/ ( 6 x 10^-3) = 2.95 x 10^-12 F

a) Energy = 0.5 cV^2= 0.5(  2.95 x 10^-12)( 5^2) = 3.6875 x 10^-11 J

b) A half filled capacitor can be thought as a combination of two capacitors ( one filled with air and other filled with dielectric)

U( air filled) = 0.5 (0A/ 2 d)v^2 ( area gets half)

U ( dielectric filled) = 0.5( K0A/ 2 d)v^2

Total U =  0.5 (0A/ 2 d)v^2 + 0.5( K0A/ 2 d)v^2 = 0(A/ 4 d)v^2 { k+1) = 25.65 x 10^-12 J

c) When the capacitor is disconnected with battery, the charge will stop chhanging , that also implie that capacitor will have the same charge calculated in part (b)

E = 1/2 Q^2/C

Q^2 = 2EC = 2 0(A/ 4 d)v^2 { k+1)   { k+1)  0 A/2d}

Q = (k+1)(0)( A)( v/ 2d) = ( 3)( 8.85×1012 ) ( 10 x 10^-4) (5)/ ( 2 x 6 x 10^-3) = 11.06 x 10^-12 C apprx

C ( filled with air) = 0 A/d = 1.475 x 10^-12 F

U = 1/2 Q^2/C = 0.5 ( 11.06 x 10^-12)^2 / 1.475 x 10^-12= 41.465 x 10^-12 J apprx

d) Work done = enrgy of capacitor when dielectrc is completely removed - enrgy of capacitor when dielectrc is half removed =  41.465 x 10^-12- 25.65 x 10^-12 = 15.815 x 10^-12 J apprx

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