FINDING ANALYTICAL SOLUTIONS TO ELETROMAGMETIC PROBLEMS
Task
In this mini project calculate the Electrical field variation, equipotential contours and plot the results to given two questions.
Question 01:
Two grounded semi-infinite metallic plates are placed on the x−axis and the y−axis of a Cartesian coordinate system to form a 90°corner. A positive charge 
is located at the point (a, a) where a is arbitrary. Plot equipotential contours surrounding this charge and the expected electric field.
Question 02:
A charge of +Q is uniformly distributed in the central region 0 ≤ r ≤ a and – Q C is distributed in the external region b ≤ r ≤ c of a coaxial cable. Find the electric filed in all regions of the coaxial cable and plot the results.
Theory
Coulomb’s Law
The electric force between two charges is proportional to the product of two charges and inversely proportional to the square of the distance between the two charges.
If two charges having different signs, they will attract each other. If two charges have same signs, they will repel each other.
Electric Field
The electrical field caused by a charge Q is a vector field that defined as,
Where, E – Electric field, F – Force between Q and q charges.
If the electric field not varying with the time, it is said as ‘Electrostatic field’. Also, the electric field in a region due to the charge Q is written as.
Also, these electric fields are start from positive charges and terminate on a negative charge. The electric field due to multiple charges, is same as the vector sum of the field due to individual components.
Charge Densities
There are 3 charge densities can define, volume charge density, Surface charge density and line charge density. Volume charge density is a amount of charge that have in a infinity small volume block. So ot can be calculate as,
If the charge density not varying with the position, it said ‘uniform volume charge distribution’. In similar way, surface charge density (𝜌𝑠), and linear charge density (𝜌𝑙) are also defined as,
Gauss Law
The total electric flux (Total electric field in a area) out of a closed surface is proportional to the total charge enclosed by the closed surface.
Electric Potential
A charge particle gaining potential energy as the particle is moved in a region against an electric field. Then, the work done will be,
The total energy that has been expended in order to bring the charges into close to a region is stored in this region as electrostatic energy and it can be recover later. The lines, surface or volume that have same potential known as equipotential contours, surface or volume. As the ohm law, it can be showing the ideal conductor is an equipotential surface. So the electric field is always perpendicular to the metal or conductor surface. To simplify the calculations, It can include Image charges as satisfy this requirement in near conductor.
Also, it can rearrange above equation as subjected to electric field,
E = −∇𝑉
Answer to Q1:
Results
For demonstration, let a = 2;
Equipotential contours and Electric field.
It act as quadrupole system.
At the metallic plates(0,0 position), the potential is zero. Electric field not is always prependicular to the grounded metallic plate and Electric field directed to metallic frame.
MATLAB Code:
s = 0.2; % spacing factor
a= 2;%defining a
[X,Y]=meshgrid(-3*a:s:3*a,-3*a:s:3*a); % Making a space
%defining the potential function
V = (1./(sqrt((X-a).^2 + (a-Y).^2)))- (1./(sqrt((X + a).^2 + (a
Y).^2)))+(1./(sqrt((X+a).^2 + (a+Y).^2)))-(1./(sqrt((X-a).^2 + (a+Y).^2)));
hold on;
contour(X,Y,V,100) %Plot the contours
Answer to Q2:
During the region that positively distributed charge, the electric field is growing up linearly with the radius. Then in free space, electric field reduce with radius inverse proportionally. In negatively charged region, the electric field is declining rapidly until become zero. Therefore, the outer space, not effected electric field due to this charges that belong in coaxial cable.
Matlab Code
clc; clear all; close all;
%Initial Parameters
Q = 1*10^-9; l=1/(2*pi*9);ep = 9*10^-9;
a = 1;b= 2; c= 3;
% defining r
r_a = 0:0.1:a ;
r_b = a+0.1:0.1:b ;
r_c = b+0.1:0.1:c ;
E_a = (Q.*r_a)./(2.*pi.*a.^2.*l.*ep); %Electric field in 0<r<a
E_b = Q./(2.*pi.*l.*ep.*r_b); % Elecrtic field a<r<b
E_c = (Q./2.*pi.*r_c.*l.*ep).*(1-(r_c.^2-b.^2)./(c.^2-b.^2)) %Electric field b<r<c
plot([r_a r_b r_c],[E_a E_b E_c]); %Plotting Electric field vs r