Electric field and a misbehaving particle

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Richard 2012-3-29

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https://ww2.mathworks.cn/matlabcentral/answers/33843-electric-field-and-a-misbehaving-particle

What I wish to do is to model the trajectory of a point particle of small negative charge in the field of a point charge of large positive mass, so the latter is stationary.

What I have done so far always gives the solution where the small particle is totally unaffected by the field it's in -- so it just travels on in a straight line, which is obviously rubbish. (It should be going in a circle centered at the origin!)

Could anyone please help me? Note that the code is in its rough/draft stage so signs and constants are not guaranteed, but that should not affect the general physics behind it!

We shall consider motion in a plane. The coordinates of the particle with small charge is (a(t),b(t)) and that with large charge is at (0,0).

function EField
clear 
trange = (0:.005:5);
%Starting conditions
a0 = ... ;
b0 = ... ;
adot0 = ... ;
bdot0 = ... ;
addot0 = -a0/((a0^2+b0^2)^(3/2));
bddot0 = -b0/((a0^2+b0^2)^(3/2));
vec0 = [a0, b0, adot0, bdot0, addot0, bddot0];
options = odeset('RelTol', 1e-8 ,'AbsTol', 1e-10);
[t,vec] = ode45(@EF, trange, vec0, options);
%Plot trajectory
plot(vec(:,1), vec(:,2));
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

This is the function EF in a separate m-file.

function sth = EF(t,vec)
a = vec(1);
b = vec(2);
adot = vec(3);
bdot = vec(4);
addot = vec(5);
bddot = vec(6);
sth = zeros(size(vec));
sth(1) = a;
sth(2) = b;
sth(3) = adot;
sth(4) = bdot;
sth(5) = -a/((a^2+b^2)^(3/2));
sth(6) = -b/(2*a^2+b^2)^(3/2));

Thank you.

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Jan 2012-3-29

What is "X" in "sth = zeros(size(X));"?!

Richard 2012-3-29

Ah, that was a typo... For some reason I couldn't copy and paste, so I retyped the code -- unfortunately not too accurately!

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Answer 1

Jan 2012-3-29

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https://ww2.mathworks.cn/matlabcentral/answers/33843-electric-field-and-a-misbehaving-particle#answer_42489

在 MATLAB Online 中打开

What about:

...
sth(1) = adot;
sth(2) = bdot;
sth(3) = addot;
sth(4) = bddot;
...

The change of the position should be related to the velocity.

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Jan 2012-3-29

I do not know to function you want to integrate. I assume, that sth(5) and (6) are correct in your original question and only the frist 4 components are wrong.

"sth" is the LHS, "vec" is the RHS of your function.

The function to be integrated is: dx/dt=... . It contains the local derivative of x. Let's use a simpler example: A point of mass 1 is accelerated in the gravitational field:

x''=g

The current state of the object is represented by x=[position, velocity]. Then the function to be integrated is:

function dx = dxdt(t, x)

dx(1) = x(2);

dx(2) = g;