Electrical Engineering Assignment: Coaxial Cable and Field Analysis

This document presents a comprehensive solution to an Electrical Engineering assignment. The solution begins by calculating the capacitance of a coaxial cable and its breakdown voltage, considering dielectric strength. It then delves into the conductivity and relative permittivity of wet soil. The assignment further explores electric fields using Gauss's law, determining the electric field in different regions of a structure, and calculating voltage differences. It also covers the total capacitance of a structure, represented by capacitors in series. The document continues to address the concepts of current density, electric fields and drift velocity using Ohm's law. Finally, it addresses concepts of solenoidal and rotational vector fields. The solution provides detailed calculations and explanations for each part of the assignment, making it a valuable resource for students studying electrical engineering.

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Question 1
Solution
(a) Capacitance of co-axial cable;
C=
2 πE
ln ( R
r ); where R=16 mm and r=7 mm
C= ( 2 π ×3.35×10-12)/ l(16/7) =67.29×10-12 F/m
So, CT= (67.29×10-12) (40×10-3) =2.6917×10-12 F
(b) Given di-electric strength=3 MV/m=Emax
So, Emax=V/(rln)(R/r)
3×106=V/ (7*10-3ln) (16/7)
V=17.36 kV
(C) Given that Rin=2Ω, C=11 PF
As we know that
Rin= T ins
2 πd l n ( R
r )

2 ×103= T ins
2 π × 40 ×10−3 ln ( 16
7 )
T ¿= 2 π × 40 ×10−3 ×2 ×103
l n ( 16
7 ) =608.04 Ω/m
So, Conductivity= 1
Tins
=1.644 ×10−3 siemen−m
(d) C=
2 π ∈0 ϵr
l n ( R
r ) ×l
11 ×10−12= 2 π × 8.854 ×10−12 ϵ r
l n (16
7 ) × 4 × 10−3
ϵr =4.08
Question 2
(a) a<r<r3; ∫ E . ds= Rln
ε
E .2 πrh= R
εr εo
=Er−s 2= Q
2 π εr εo rh
For r3<r<b
E .2 πrd= Q
εo
=Er−g 2= Q
2 π εo rh
(b) V a =−∫
∞
r 3
E . dr=∫
b
r 3
Eg 2 . dr −∫
b
a
Eg 2 . dr

V a =−∫
b
r 3
Q
2 π εo rh −∫
rs
a
Q. dr
2 π εr ε o rh
V a = −Q
2 π εo d ln ( r s
b )− Q . dr
2 π εr εo rh
V rs=−∫
b
rs
Eg 2 .dr =−¿ Q
2 π εo d ln ( rs
b )¿
V a −V rs= Q
2 π εo rh ln ( rs
b )− Q . dr
2 π εr εo rh ln ( a
r )
V b =0
V g 2= Q
2 π εo d ln ( rs
b )
Total V=V s 2 +V g 2
V = Q
2 π εo
[ ln ( r s
2 )
εr h +
ln ( rs
b )
l ]
(c) The capacitor is in series with each other
Cg 2= Q
V g 2
= 2 π ε o d
ln ( rs
b )
Cs 2= Q
V s 2
=2 π εo d
ln ( rs
a )
(d) Resistance, R= 1
σs
h
2 π r s h

R= 1
2 π rs σ s
Question 3

Question 4
A=Area of the cross section=width*thickness=0.1*1*10-6
Isigma=conductivity of the copper=5.76*102
So we find EQ=10-3 A/ (10-2*5.76*102) =1.736*10-3 N/m
(c) The drift velocity is the product of the mobility and the electric field

Vd=ImuEq=3.2*10-3*1.736*10-3=555.55*10-9
(d) Voltage difference between the top and bottom edge of copper strip
=B (l*V)
1(w*V) volt
(e) Voltage gain of the system per guays = (w*u)*1000
Question 4
Given data
Ohm’s law states that the current of any circuit is directly proportional to the voltage across the
point and the constant is nothing of the resistance between them. Now we can derived and
discuss the point form of this law which is basically the relation between the current density and
the conductivity and the electric field. Before the magnetic force is calculated the vital magnetic
force which is the force developed due to magnetic force actually the older name of this is called
Lorenz force
This force can be calculated as forced can be represented as follows
Fm= (q*4).B
Eh=q*4.B/q
(b) From the above shown derivation we get the electric field as follows:
J=sigma*Ea
Ea=J/sigma=J/ (A*sigma)

I=current across the circuit=10-3 A
Question 5
A vector field A is said to be solenoidal if
Such a field has neither nether source nor sink of flux
A vector field A is said to be rotational or potential if
Thus in an isolational field A, the circulation of A around a close path is identically zero
An isolational filed is also known as conservative field
Given the vector field as