"Index exceeds the number of array elements"

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Hello,
I am recieving this error:
Index exceeds the number of array elements (1).
Error in sym/subsref (line 890)
R_tilde = builtin('subsref',L_tilde,Idx);
Error in E_field (line 7)
Ey = matlabFunction(E(2));
Here is the function file the error is referencing:
function [Ex, Ey, Ez] = E_field()
syms x y z
R_s = 0.02;
V = 0;
epnaut = 8.854187E-12;
k = 1/(4*pi*epnaut);
Q = (V*R_s)/k;
r = [x, y, z];
E = (k*Q/norm(r)^3)*r;
Ex = matlabFunction(E(1));
Ey = matlabFunction(E(2));
Ez = matlabFunction(E(3));
end
  1 Comment
Walter Roberson
Walter Roberson on 29 Jan 2020
The posted code works for me. However you should using the 'vars' option of matlabFunction or else your function handle is not going to accept any arguments because your R values are all constants.

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Accepted Answer

Matt J
Matt J on 29 Jan 2020
Edited: Matt J on 29 Jan 2020
Perhaps this is what you are looking for,
function [Ex, Ey, Ez] = E_field(x,y,z)
R_s = 0.02;
V = 0;
epnaut = 8.854187E-12;
k = 1/(4*pi*epnaut);
Q = (V*R_s)/k;
r = [x, y, z];
E = (k*Q/norm(r)^3)*r;
Ex = E(1);
Ey = E(2);
Ez = E(3);
end
  9 Comments
Walter Roberson
Walter Roberson on 1 Feb 2020
Is there a reason you are using global instead of persistent ? Is there some other function that will use the created Bx By Bz Ex Ey Ez ?
Tom Keaton
Tom Keaton on 4 Feb 2020
Sorry I missed this question. The answer is yes which is why I am using it. I will show the code that causes this issue in another post.

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More Answers (1)

Walter Roberson
Walter Roberson on 1 Feb 2020
V = 0;
V is 0.
Q = (V*R_s)/k;
V is 0, so Q will be 0.
E = (k*Q/norm(r)^3)*r;
Provided that norm(r ) is not 0 and k is not infinity, then because Q is 0, E is going to be
zeros(size(r))
symvar() of a vector of 0 is empty. matlabFunction() of a constant, when you do not use any 'vars' option, is a function handle that accepts no inputs and returns a scalar copy of the constant.
The original code that was posted does not give me an index exceeded matrix dimension when I test it in R2019a.
  1 Comment
Tom Keaton
Tom Keaton on 4 Feb 2020
Under this specific circumstance this does work for me now.

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