a. Derive the capacitance of a cylindrical capacitor with inner plate radius w1,

Question

a. Derive the capacitance of a cylindrical capacitor with inner plate radius w1, outer plate radius w2, length z, and dielectric constant kappa and determine how the capacitance is affected if the dielectric constant is doubled while the length is halved.

b. A parallel-plate capacitor contains two layers of equal thickness of dielectric materials. One has a permittivity four times that of empty space. The other has twelve times that of empty space. The capacitor is charged to 120 V and disconnected from the battery. The second layer (12 epsilon-naught) is ripped out, leaving air in its place. What is the new voltage on the capacitor?

Explanation / Answer

a)

inner plate radius =w1, outer plate radius =w2, length =z, dielectric constant k

Ccylinder= 2k0z/ln(w2/w1)

Now when k’=2k and z’=z/2

Ccylinder‘ = [2k’0z’]/ln(w2/w1) = [2*2k0z/2]/ln(w2/w1) = 2k0z/ln(w2/w1)

Thus,

Ccylinder‘ = Ccylinder = 2k0z/ln(w2/w1)

b)

C = k0A/d

Initial condition,

C1i = k1i0A/d = 40A/d=4C

C2i = k2f0A/d = 120A/d=12C

C1 and C2 are in series hence

1/Ci = 1/C1 + 1/C2 = 1/(4C)+1/(12C)

Ci = (4*12)C/(4+12)= 3C

Qi = CiVi = (3C)*120 = 360C --------------(1)

Final condition,

C1f = k10A/d = 0A/d = C

C2f = k2i0A/d = 120A/d = 12

C1f and C2f are parallel hence

1/Cf = 1/C1f + 1/C2f = 1/(C)+1/(12C)

Cf = (1*12)C/(4+12)= 0.75C

Qf = CfVf

Qf= CfVf = (C)*Vf = 0.75C*Vf --------------(2)

Special condition, Qf= Qi

0.75C*Vf = 360C

Vf = 360/0.75

Vf = 480 V

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