ELEC2015
Electromagnetic Applications – Part B
School of Electrical Engineering & Telecommunications
ELEC2015 - Electromagnetic Applications
PART B – Low Frequency Applications
Tutorial 1 – Application of Gauss’s Law
1
The parallel-plate arrangement of figure Q1 may have two following dielectric materials
between them:
(i) air:
(ii) mica:
εr = 1
εr = 6
10 cms
20 cms
0.5 mm
Figure Q1
The left and the right plates have charges of +0.4 and –0.4 micro-Coulombs respectively.
(a) What assumptions must you have in order to use Gauss’s law to find the electric field
in the space between the plates? For each dielectric,
(b) find the electric field E and the electric flux density D vectors between the plates
using Gauss’s law. How would this field vary with separation between the plates?
(c) Find the potential difference V between the plates. How would the potential difference
vary with separation between the plates?
[Ans: Eair = 2.259 MV/m, Emica = 0.3765 MV/m, Vair = 1130 V, Vmica = 188 V]
2.
The two parallel plates of figure Q1 are connected to a dc voltage source of 800 volts. The
left side plate is at the higher potential. For the two dielectric materials of question 1,
(a) find the electric field E and the electric flux density D vectors between the plates
using Gauss’s law. How would this field vary with separation between the plates?
G G
G
G
G
G
[ E = an 1.6 × 10 6 V / m,Dair = an 14.167 × 10 −6 C / m2 ,Dmica = an 85.2 × 10 −6 C / m2 ]
(b) Find the charge Q on each plate. How would the Q vary with separation between the
plates?
[ Qair = 0.283 × 10 −6 C / m2 ,Qmica = 1.7 × 10 −6 C / m2 ]
Tutorial 1 – Application of Gauss’s Law
1
F. Rahman/September 2003
ELEC2015
3.
Electromagnetic Applications – Part B
For the parallel plate arrangement of Q1, and the charges of 0.4 µC on each plates,
(a) find the capacitance of the plates for the two dielectric materials.
[ Cair = 3.54 × 10−10 F (i.e., 0.354 nF ) , Cmica = 21.24 × 10−10 F , (i.e., 2.14 nF ) ]
(b) How does the capacitance of the arrangement vary with separation between the
plates?
4.
A parallel-plate capacitor of 0.5 m2 area and plate separation of 30 mm is filled with two
stacked dielectric slabs. The upper slab thickness is 20 mm with εr = 6 and the lower slab
thickness of 10 mm with εr = 12. If a potential of 200V is applied to the capacitor, find the
values of the following quantities in each slab: (a) electric field E, (b) flux density D, (c)
polarisation vector P, and (d) the total capacitance of the capacitor. Assume no fringing of
flux.
[Ans: E = 4 & 8 kV/m, D = 425 nC/m2, P = 354 & 390 nC/m2, C = 1.06 nF]
5.
A uniform dielectric medium (εr = 9) of large extent has a uniform flux density D = 15
pC/m2 applied. (a) Find D inside a thin disc-shaped cavity cut in the dielectric with flat
sides normal to D. (b) Find D inside a needle-shaped cavity with axis parallel to D. The
cavities have air inside.
[Ans: (a) Dair = 15x10−12 C/m2, (b) Dair = 1.67x10−12 C/m2].
6.
A potential of 50 kV RMS is applied between a pair of parallel-plate electrodes of
separation 2cm, with air as the dielectric. If the dielectric strength of air is 3 MV/m,
investigate whether the insertion of a sheet of mica of thickness 5mm between the
electrodes will provide additional protection against breakdown, given that mica is a far
superior insulator than air. The dielectric strength of mica is 10 MV/m and its relative
permittivity εr is 4.
7.
A single-core power cable of circular cross section, as indicated in the figure Q7, has a
single insulation layer of polystyrene of dielectric strength 20MV/m. The relative
permittivity of polystyrene is 2.5. The copper cable has a diameter of 0.8 cm. The cable is
surrounded by air, the dielectric strength of which is 3 MV/m. Assume that the electric
fields within the dielectric and in the air are not to exceed half their breakdown strengths.
(a) Calculate the required thickness of the insulation for the cable and its maximum
working voltage. Assume that the outside of the cable insulation is to be at ground
potential.
[6.27 cm, 112.5 kV]
(b) Plot the variation of E and V in the conductor and the dielectric layer.
(c) Discuss the inverse problem where the maximum operating voltage may be known
and the insulation (dielectric) material and its thickness have to be determined.
Polystyrene
εr = 2.5
Emax = 20MV/m
d = 0.8 cm
Figure Q7
Tutorial 1 – Application of Gauss’s Law
2
F. Rahman/September 2003
ELEC2015
8.
Electromagnetic Applications – Part B
A coaxial cable as shown in figure Q8 is used to carry electrical power. The radius of the
inner conductor is 0.4 cm. It is covered with two concentric dielectric layers of rubber (εr
= 3.2) and polystyrene (εr = 2.6). The inner radius of the polystyrene layer is rp. The inner
radius of the outer conductor layer, ro, is 0.832 cm.
(a) Assuming that maximum electric field intensities in the dielectric (insulating)
materials do not exceed 25% of their breakdown strengths, determine the voltage
rating for the of the cable for
(i)
rp = 1.75ri
and (ii)
rp = 1.35ri
The breakdown strengths of the two dielectric materials are:
Rubber
Polystyrene
25 MV/m
20 MV/m
εrp = 2.6
εrr = 3.2
ri
rp
ro
Figure Q8
(b) Plot the variation of E and V in the dielectrics and the conductor.
9.
A coaxial capacitor of length L (figure Q9) has inner conductor of radius of 1 cm, and
outer conductor has inside radius of 9 cm. It is filled with three dielectrics of relative
permittivities εr = 8 from radius r = 1 to 4 cm, εr = 1 from r = 4 to 5 cm and εr = 20 from
r = 5 to 9 cm.
(a) Find
(i) the capacitance of the capacitor/meter.
[0.13 nF/m]
(ii) its breakdown voltage if the dielectric strengths of the materials with εr = 8 and εr
= 20 are 20 MV/m and 40 MV/m respectively. The dielectric with εr = 1 is air,
with a dielectric strength 3 MV/m.
[51.4 kV]
Tutorial 1 – Application of Gauss’s Law
3
F. Rahman/September 2003
ELEC2015
Electromagnetic Applications – Part B
(b) Calculate the electric fields in the three layers just before breakdown.
[1.5 MV/m, 3 MV/m, 0.118MV/m]
(c) Will any of the other two layers also breakdown when the maximum allowable field in
the weakest layer is just exceeded? Calculate electric field in the other two layers
when the weakest layer has just broken down.
[No, 3.12 MV/m, 0.25 MV/m]
(d) Plot the variation of E and V in the dielectrics.
εr = 20
εr = 1
εr = 8
1
4
5
cm
9
Figure Q9
10.
If the breakdown dielectric strength of air is 3 MV/m, estimate the maximum voltage that
can be used, without breakdown, between two parallel overhead lines forming a two-wire
transmission line. The two lines are each of bare, smooth-surface conductors of diameter
10 mm. The separation between the conductors is 1.0 meter. Assume that the ground
beneath has no effect.
[159 kV peak]
11.
A constant voltage of 400V is applied to the parallel plate capacitor shown in the figure
Q1.
(a) Calculate the force between the plates.
(b) How will this force vary with voltage and separation?
12.
[ F = −0.057a x N]
Calculate the force between the two parallel conductors of problem 10, when operating at
the maximum allowable voltage.
[ F = −0.013ar N]
Tutorial 1 – Application of Gauss’s Law
4
F. Rahman/September 2003