Field Line Plot

There are two uniform spherical charge distributions, each with radius 1 1 . Distribution 1 1 is centered at ( x 1 , y 1 , z 1 ) = ( 2 , 0 , 0 ) (x_1, y_1, z_1) = (-2,0,0) and has charge + 1 +1 . Distribution 2 2 is centered at ( x 2 , y 2 , z 2 ) = ( 2 , 0 , 0 ) (x_2, y_2, z_2) = (2,0,0) and has charge 1 -1 . Let us plot the electric field lines in the following manner.

1) Begin at point ( x 0 , y 0 , z 0 ) = ( x 1 + cos θ 0 , y 1 + sin θ 0 , 0 ) (x_0, y_0, z_0) = (x_1 + \cos \theta_0, y_1 + \sin \theta_0, 0)
2) Update the coordinates as follows:

x k = x k 1 + ϵ E ^ x k 1 y k = y k 1 + ϵ E ^ y k 1 z k = 0 x_k = x_{k-1} + \epsilon \, \hat{E}_{x \, k-1} \\ y_k = y_{k-1} + \epsilon \, \hat{E}_{y \, k-1} \\ z_k = 0

3) Continue plotting until the field line hits the boundary of charge distribution 2 2
4) Make plots for the following values of θ 0 \theta_0 , resulting in 25 25 field lines: ± 12 0 , ± 11 0 , ± 10 0 , ± 9 0 , ± 8 0 , ± 7 0 , ± 6 0 , ± 5 0 , ± 4 0 , ± 3 0 , ± 2 0 , ± 1 0 , 0 \pm 120^\circ, \pm 110^\circ, \pm 100^\circ, \pm 90^\circ, \pm 80^\circ, \pm 70^\circ, \pm 60^\circ, \pm 50^\circ, \pm 40^\circ, \pm 30^\circ, \pm 20^\circ, \pm 10^\circ, 0^\circ

In step 2 2 , subscript k k denotes the present value, and subscript ( k 1 ) (k-1) denotes the previous value. E ^ \hat{E} is a unit-length version of the net electric field at point ( x , y , z ) (x,y,z) . The parameter ϵ \epsilon is a very small number.

What is the combined length of all 25 25 field lines?

Details and Assumptions:
1) 1 4 π ϵ 0 = 1 \frac{1}{4 \pi \epsilon_0} = 1
2) To check results, it helps to run the plotter with ϵ \epsilon values spanning several orders of magnitude.


The answer is 150.6.

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1 solution

Karan Chatrath
Feb 27, 2020

Nice problem as always. This time, I am only pasting the script of code I wrote to solve this. I hope it is readable. I will make amends to better it, if necessary.

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clear all
clc

% Centers of spheres:
r1 = [-2;0;0];
r2 = [2;0;0];

% Initialisation of epsilon and total length:
eps = 0.00001;
L = 0;

% Varying theta from -120 degrees to 120 degrees in steps of 10 degrees:
for theta = -120*(pi/180):10*(pi/180):120*(pi/180)

    % Initial point for a field line:
    Ro = r1 + [cos(theta);sin(theta);0];

    % Final point for a field line:
    R1 = r2 + [cos(pi - theta);sin(pi - theta);0];

    % Initialisation of the vector of coordinates of a point unit charge
    % moving due to the influence of the combined field due to each sphere:
    x(:,1) = Ro;

    % Loop running till test point charge moves 
    % from left sphere to the right:
    k = 2;
    while norm(x(:,k-1)-R1) >= 1e-3

        % Calculation of electric field as per Coulomb's Law:
        % Note that x is a vector having 3 components:
        RL = x(:,k-1) - r1;
        RR = x(:,k-1) - r2;
        E  = RL/((norm(RL))^3) - RR/((norm(RR))^3);

        % Unit vector of net electric field:
        E_hat = E/norm(E);

        % Trajectory calculation as per instruction in problem statement:
        x(:,k) = x(:,k-1) + eps*E_hat;

        % Combined length calculation
        % Euclidean length between two successive points on a field line:
        L = L + norm(x(:,k)-x(:,k-1));

        k = k+1;

    end
end

% Answer: 150.5695

Thanks for the solution. I would also like to make the plots for both distributions having positive charge, but the "while" loop conditions might be a bit trickier to establish.

Steven Chase - 1 year, 3 months ago

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You could create a boundary which is a larger circle (containing the charge distributions) centred at the origin. So that's where the lines could be plotted till.

Karan Chatrath - 1 year, 3 months ago

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@Karan Chatrath Sir I always like Your solution . Can you please post solution Of this question https://brilliant.org/problems/hemisphere-e-field-quantitative/. In your style . First taking position vector . Actually I am not able to take elemental area and all those things can you help me?

A Former Brilliant Member - 1 year, 3 months ago

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@A Former Brilliant Member I will post when I can. In the meantime, ask google about 'the derivation of a surface area element in spherical coordinates'. You should find some helpful material. Also, recall that an area vector is always normal to a surface.

Karan Chatrath - 1 year, 3 months ago

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