How can I find the electrical field at (0.56, 0.70)?

Find analytical value of charge on the

    % coaxial cable capacitor plates
    Q = 10*2*pi*L / (log(0.8) - log(0.2))
    % find analytical values of the electric field in the
    % dielectric
    E = Q ./(2*pi*vr*L)
    El = (s(2:end,1) - s(1:end-1,1))/hr
    
    % Find numerical values of electric field (the s contains values of
    % electric potential)
    En = (s(2,:) - s(1,:))/hr
    
    % Just check the integral of the charge (should match the analytical
    % value)
    Q = sum(En(1:end-1))*hz
    
    % find value of the capacitance
    C = Q/U
    
    % This should produce the voltage U, as this represents the
    % integral over r from r_a to r_b (Stokes Theorem)
    sum(El)*hr

[Image Analyst]

What is L, vr, s, hz, etc.?

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[suketu vaidya]

U=2;   
L = 1; % total height of the coaxial cable
    hr = 0.1;
    hz = 0.1; % these should be the r coordinates of grid
    vr = 0.2:hr:0.8;
    vz = 0.2:hz:0.8;

[suketu vaidya]

and s is my 2d boundry condition problem for The electric field in a coaxial cable it its 2d graph of

[suketu vaidya]

Answer 1

[Kathryn Akmal]

You can use the find() command to find the index of the x and y values you're interested in, and take that index of the array containing the electrical field values. Something like this:

xindex=find(0.56);
yindex=find(0.70);
EFieldValue=EField(xindex,yindex);