Capacitor with a Dielectric Problems and Solutions

 Problem #1

An air-filled parallel-plate capacitor has a capacitance of 1.3 pF. The separation of the plates is doubled, and wax is inserted between them. The new capacitance is 2.6 pF. Find the dielectric constant of the wax.

Answer;
Known:
Capacitance an air-filled parallel-plate capacitor is C₀ = 1.3 pF = 1.3 x 10^-12 F
Capacitance an wax-filled parallel-plate capacitor is C₀ = 1.3 pF = 2.6 x 10^-12 F, then

 C₀ = ε₀A/d = 1.3 pF (*)

where: ε₀ = permittivity of vacuum/air, A = area of each plate of the capacitor, d = distance of separation between the plates of the capacitor

Capacitance of the wax-filled capacitor

C = ε₀εA/(2d) = 2.6 pF    (**)

ε = dielectric constant of wax

Dividing (**) by (*), we have:

C/C₀ = 2.6/1.3 = 2 = ε/2

so, ε = 4

Problem #2
A coaxial cable used in a transmission line has an inner radius of 0.10 mm and an outer radius of 0.60 mm. Calculate the capacitance per meter for the cable. Assume that the space between the conductors is filled with polystyrene.

Answer;
Known:
inner radius, a = 0.10 mm
outer radius, b = 0.60 mm

the capacitance for the cable given by

C = 2πLε_rε₀/[ln (b/a)]

With ε_r = permittivity relative material

Then,

the capacitance per meter for the cable is

C/L =  2πε_rε₀/[ln (b/a)]

       = 2π(2.6)(8.85 x 10^-12 C²/Nm²)/[ln (0.60/0.10)] = 81 x 10^-12 F/m

C/L = 81 pF/m

Problem #3
A parallel-plate air-filled capacitor has a capacitance of 50 pF. (a) If each of its plates has an area of 0.35 m², what is the separation? (b) If the region between the plates is now filled with material having ε = 5.6, what is the capacitance?

Answer;
Known:
Capacitance an air-filled parallel-plate capacitor is C₀ = 50 pF = 5.0 x 10^-11 F
Area plates, A = 0.25 m²

(a) the separation plates given by

C = ε₀A/d or

d = ε₀A/C = (8.85 x 10^-12 C²/Nm²)(0.25 m²)/(5.0 x 10^-11 F)

d = 0.04425 m = 44.25 mm

(b) If the region between the plates is now filled with material having ε = 5.6, the capacitance is

C =  ε₀εA/d = εC₀ = 5.6 x 50 pF = 280 pF

Problem #4
You are asked to construct a capacitor having a capacitance near 1 nF and a breakdown potential in excess of 10000 V. You think of using the sides of a tall Pyrex drinking glass as a dielectric, lining the inside and outside curved surfaces with aluminum foil to act as the plates. The glass is 15 cm tall with an inner radius of 3.6 cm and an outer radius of 3.8 cm. What are the (a) capacitance and (b) breakdown potential of this capacitor?

Answer;
Known:
The capacitance of a cylindrical capacitor of length L = 0.15 m,
Inner radius, a = 0.036 m
Outer radius, b = 0.038 m
and dielectric material with relative permitivity ε_r = 4.7 (for pyrex glass)

(a) Capacitance is given by

C =  2πLε_rε₀/[ln (b/a)]

   = 2π x 8.85 x 10^-12 C²/Nm² x 4.7 x 0.15 m/[ln (3.8/3.6)]

C = 0.73 x 10^-9 F = 0.73 nF

(b) the breakdown potential is given by

V = dielectric strength x thickness

For pyrex the electric strength is 14 kV/mm

The thickness of the capacitor is 0.038 – 0.036 = 0.002 m = 2 mm, then

V = 14 kV/mm x 2 mm = 28 kV