NaN at (5,0) in graph. How to change it? Electric field

As you can see, at the coordinates (5,0) it has NaN NaN

Does anyone know what can i do?

x = -24:18;

y = (-20:2:18).';

E0 = 8.85e-12;

k = 1/(4*pi*E0);

q = 2;

aq = abs(q);

a = 10;

Er_x = 0;

Er_y = 0;

id = -4;

jd = -12;

while id < 4

xp = ((x + a/2).^2 + (y-id).^2).^1.5;

E1_x = k * aq * (x+a/2) ./ xp;

E1_y = k * aq * y ./ xp;

Er_x = Er_x + (E1_x);

Er_y = Er_y + (E1_y);

id=id+1;

end

while jd < 12

xm = ((x - a/2).^2 + (y-jd).^2).^1.5;

E2_x = -k * aq * (x-a/2) ./ xm;

E2_y = -k * aq * y ./ xm;

Er_x = Er_x + ( E2_x);

Er_y = Er_y + ( E2_y);

jd=jd+1;

end

Er_x = Er_x + (E1_x + E2_x);

Er_y = Er_y + (E1_y + E2_y);

Etot = sqrt(Er_x.^2 + Er_y.^2);

u = Er_x ./ Etot;

v = Er_y ./ Etot;

quiver(x,y,u,v)

Answer 1

[KSSV]

You can fill the NaN's using fillmissing Read about this function.

x = -24:18;
y = (-20:2:18).';
E0 = 8.85e-12;
k = 1/(4*pi*E0);
q = 2;
aq = abs(q);
a = 10;
Er_x = 0;
Er_y = 0;
id = -4;
jd = -12;
while id < 4
    xp = ((x + a/2).^2 + (y-id).^2).^1.5;
    E1_x =  k * aq * (x+a/2)  ./ xp;
    E1_y =  k * aq * y        ./ xp;
    Er_x =  Er_x + (E1_x);
    Er_y =  Er_y + (E1_y);
    id=id+1;
end
while jd < 12
    xm = ((x - a/2).^2 + (y-jd).^2).^1.5;
    E2_x = -k * aq * (x-a/2)  ./ xm;
    E2_y = -k * aq * y        ./ xm;
    Er_x =  Er_x + ( E2_x);
    Er_y =  Er_y + ( E2_y);
    jd=jd+1;
end
Er_x =  Er_x + (E1_x + E2_x);
Er_y =  Er_y + (E1_y + E2_y);
Etot = sqrt(Er_x.^2 + Er_y.^2);
u = Er_x ./ Etot;
v = Er_y ./ Etot;
% Fill the NaN's
u = fillmissing(u,'linear') ;
v = fillmissing(v,'linear') ;

quiver(x,y,u,v)

[Cristobal Meza]

oh I see, but the thing is that:

I need not to have a value, let me explain, it has to remain like the other ones above or below.

lets say that there are two graphs, the one in the left is perfect.

The one in the right has exactly filled (with NaN) at coordinates (5,0) but it shouldn`t

The only thing that i need to change is exactly at (5,0)

However, it's very interesting to know that there is such function, thank you for that.

[Walter Roberson]

The calculation you posted leads to 0 divided by 0 at that location. What should the result be there?

Answer 2

[Walter Roberson]

Etot = sqrt(Er_x.^2 + Er_y.^2);
u = Er_x ./ Etot;

What happens if Etot is 0 because Er_x and Er_y are both 0? 0/0 is... what?

Hint:

You can change Etot to the maximum of the existing calculations together with a very very small value. Then you would have 0/tiny which would be 0.