Matlab code for Engineering Physics not working
显示 更早的评论
CODE
%Assigning values of required constants
k=1/(4*pi*8.542e-12) ;
z=linspace(-0.5,0.5) ;
E=zeros(size(z));
Q=1e-10;
%Creating a matrix containing position vectors for the 100 points on the z-axis
rz=linspace(-0.5,0.5) ;
rx=zeros(size(rz)) ;
ry=zeros(size(rz)) ;
r=[rx ; ry ; rz] ;
%Assigning variables to the position vectors
syms x [100 1]
for i=1 : 100
x(i)=transpose(r(: , i)) ;
end
%Creating a matrix containing position vectors for the 100 points on the y-axis
r1y=linspace(-0.5,0.5) ;
r1z=zeros(size(r1y)) ;
r1x=zeros(size(r1y)) ;
r1=[r1x ; r1y ; r1z] ;
%Assigning variables to the position vectors
syms x1 [100 1]
for i=1 : 100
x1(i)=transpose(r1(: , i)) ;
end
%Calculating distance between the points on z-axis and y-axis
for m=1 : 100
xi(m)=norm(x(m) - x1(m)) ;
end
E=zeros(z) ;
for m=1:100
E(m)=k*Q / abs(xi(m)) ^ 2 ;
end
%Analytical calculation and plotting graph of the solution
cons=1/(2*pi*8.542e-12)*1e-10 ;
y(1,1:100) = cons./abs(z) ;
plot(z,y,"red") ;
%Plotting graph of the solution obtained through code
hold on ;
box on ;
grid on ;
plot(z,E,"*") ;
xlabel('Distance h') ;
ylabel('Magnitude of E') ;
ERROR
Unable to perform assignment because the left and right sides have a different number of elements in assigning position vectors part.
THIS IS MY CODE PLEASE SUGGEST POSSIBLE CORRECTIONS
2 个评论
Alan Weiss
2021-12-9
What errors are you getting? What is the specific issue? Please indicate exactly what you need help with.
Alan Weiss
MATLAB mathematical toolbox documentation
Dyuman Joshi
2021-12-9
Extending Alan's comment - Your syntax is wrong in many places, not sure if that's because you have copy-pasted your code. Correct it and then mention what problems you are facing.
采纳的回答
rz=linspace(-0.5,0.5) ;
rx=zeros(size(rz)) ;
ry=zeros(size(rz)) ;
r=[rx ; ry ; rz] ;
%Assigning variables to the position vectors
syms x [100 1]
size(r)
for i=1 : 100
x(i)=transpose(r(: , i)) ;
end
So r is 3 x 100. You loop through by columns, extracting a 3 x 1 vector from it. You transpose that 3 x 1 vector to get a 1 x 3 vector. That is your right hand side. Then you try to assign that 1 x 3 vector to the scalar output location x(i) .
In MATLAB, it is not possible to define a numeric collection with () style indexing such that by using a single subscript, you get the content of the collection. It is possible to do that using {} style indexing.
2 个评论
%Assigning values of required constants
k=1/(4*pi*8.542e-12) ;
z=linspace(-0.5,0.5) ;
E=zeros(size(z));
Q=1e-10;
%Creating a matrix containing position vectors for the 100 points on the z-axis
rz=linspace(-0.5,0.5) ;
rx=zeros(size(rz)) ;
ry=zeros(size(rz)) ;
r=[rx ; ry ; rz] ;
%Assigning variables to the position vectors
for i=1 : 100
x{i}=transpose(r(: , i)) ;
end
%Creating a matrix containing position vectors for the 100 points on the y-axis
r1y=linspace(-0.5,0.5) ;
r1z=zeros(size(r1y)) ;
r1x=zeros(size(r1y)) ;
r1=[r1x ; r1y ; r1z] ;
%Assigning variables to the position vectors
for i=1 : 100
x1{i}=transpose(r1(: , i)) ;
end
%Calculating distance between the points on z-axis and y-axis
for m=1 : 100
xi(m)=norm(x{m} - x1{m}) ;
end
E=zeros(size(z)) ;
for m=1:100
E(m)=k*Q / abs(xi(m)) ^ 2 ;
end
%Analytical calculation and plotting graph of the solution
cons=1/(2*pi*8.542e-12)*1e-10 ;
y(1,1:100) = cons./abs(z) ;
plot(z,y,"red") ;
%Plotting graph of the solution obtained through code
hold on ;
box on ;
grid on ;
plot(z,E,"*") ;
xlabel('Distance h') ;
ylabel('Magnitude of E') ;
Academician
2021-12-10
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Choose a Solver 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
