FE Electrical and Computer Electromagnetics Practice Problems

The current FE Electrical and Computer specification assigns 4-6 of the 110 exam questions to Electromagnetics. Form A contains 5 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A coaxial feeder 30 m long has an inner conductor of radius 1.2 mm and an outer conductor whose inner radius is 6.0 mm. The intervening space is filled with polyethylene of relative permittivity 2.25. The capacitance of the feeder is most nearly:

(A)77.8 pF
(B)1040 pF
(C)1630 pF
(D)2330 pF

Show worked solution

Answer: (D)

The 30 m run comes to about 2330 pF, set by the logarithm of the conductor radius ratio and the polyethylene permittivity.

Charge on the inner conductor produces a purely radial field, and integrating that field from inner to outer radius yields the logarithmic potential difference behind the coaxial capacitance expression.

C = \frac{2 π ε _{0} ε _{r} ℓ}{ln ( b / a )}

Only the ratio of the radii enters, so the millimetre units require no conversion.

\frac{b}{a} = \frac{6.0}{1.2} = 5.0

ln 5.0 = 1.609

Polyethylene multiplies the air-filled value by its relative permittivity, and the 30 m length scales it linearly.

C = \frac{2 π ( 8.854 \times 1 0 ^{- 12} ) ( 2.25 ) ( 30 )}{1.609} = 2.33 \times 1 0^{- 9} F

FE Reference Handbook — Electrical and Computer Engineering: Capacitors and Inductors, Capacitance of a Coaxial Cable

Why the other choices appear

(A)Stopping at the per-unit-length value of 77.8 pF/m and never multiplying by the 30 m run.
(B)Omitting the relative permittivity of the polyethylene gives 2333/2.25 = 1037 pF.
(C)Reading 1.2 mm as a diameter makes $b / a = 10$ and $ln 10 = 2.303$ , giving 1631 pF.