082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

448 Chapter 13 Integration
Engineeringapplication13.4
Capacitanceofacoaxialcable
A coaxial cable has an inner conductor with a diameter of 1.02 mm and an outer
conductorwithaninternaldiameterof3mm,asshowninFigure13.15.Theinsulator
separatingthetwoconductorshasarelativepermittivityof1.55.Letuscalculatethe
capacitance ofthe cable per metre length.
Outer conductor
Cable sheath
r
a
b
1.02 mm 3 mm
Insulator
Inner conductor
Figure13.15
Cross-sectionofacoaxial
cable.
Before solving this problem it is instructive to derive the formula for the capacitance
of a coaxial cable. Imagine that the inner conductor has a charge of +Q per
metrelength and thatthe outer conductor has a charge of −Qper metrelength. Furtherassumethecableislongandacentralsectionisbeinganalysedinorderthatend
effects can be ignored.
Consideranimaginarycylindricalsurface,radiusrandlengthl,withintheinsulator(ordielectric).Gauss’stheoremstatesthattheelectricfluxoutofanyclosedsurfaceisequaltothechargeenclosedbythesurface.Inthiscase,becauseofsymmetry,
theelectricfluxpointsradiallyoutwardsandsonofluxisdirectedthroughtheends
of the imaginary cylinder; that is, end effects can be neglected. The curved surface
area of the cylinder is2πrl. Therefore, using Gauss’stheorem
D×2πrl=Ql
whereD =electricflux density.
WhenaninsulatorordielectricispresentthenD = ε r
ε 0
E,whereE istheelectric
field strength, ε r
is the relative permittivity, ε 0
is the permittivity of free space and
has a value of 8.85 ×10 −12 Fm −1 . Therefore,
thatis,
ε r
ε 0
E2πrl =Ql
E =
Q
2πε r
ε 0
r
(13.9)
This equation gives a value for the electric field within the dielectric. In order to
calculatethecapacitanceofthecableitisnecessarytocalculatethevoltagedifference
between thetwoconductors. LetV a
representthevoltageoftheinnerconductor and
V b
the voltage of the outer conductor.