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?

Answer 1

a)

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

C_cylinder= 2kπɛ₀z/ln(w2/w1)

Now when k’=2k and z’=z/2

C_cylinder‘ = [2k’πɛ₀z’]/ln(w2/w1) = [2*2kπɛ₀z/2]/ln(w2/w1) = 2kπɛ₀z/ln(w2/w1)

Thus,

C_cylinder‘ = C_cylinder = 2kπɛ₀z/ln(w2/w1)

b)

C = kɛ₀A/d

Initial condition,

C_1i = k_1iɛ₀A/d = 4ɛ₀A/d=4C

C_2i = k_2fɛ₀A/d = 12ɛ₀A/d=12C

C₁ and C₂ are in series hence

1/C_i = 1/C₁ + 1/C₂ = 1/(4C)+1/(12C)

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

Q_i = C_iV_i = (3C)*120 = 360C --------------(1)

Final condition,

C_1f = k₁ɛ₀A/d = ɛ₀A/d = C

C_2f = k_2iɛ₀A/d = 12ɛ₀A/d = 12

C_1f and C_2f are parallel hence

1/C_f = 1/C_1f + 1/C_2f = 1/(C)+1/(12C)

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

Q_f = C_fV_f

Q_f= C_fV_f = (C)*V_f = 0.75C*V_f --------------(2)

Special condition, Q_f= Q_i

0.75C*V_f = 360C

V_f = 360/0.75

V_f = 480 V