There are two uniform spherical charge distributions, each with radius 1 . Distribution 1 is centered at ( x 1 , y 1 , z 1 ) = ( − 2 , 0 , 0 ) and has charge + 1 . Distribution 2 is centered at ( x 2 , y 2 , z 2 ) = ( 2 , 0 , 0 ) and has charge − 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
)
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
3)
Continue plotting until the field line hits the boundary of charge distribution
2
4)
Make plots for the following values of
θ
0
, resulting in
2
5
field lines:
±
1
2
0
∘
,
±
1
1
0
∘
,
±
1
0
0
∘
,
±
9
0
∘
,
±
8
0
∘
,
±
7
0
∘
,
±
6
0
∘
,
±
5
0
∘
,
±
4
0
∘
,
±
3
0
∘
,
±
2
0
∘
,
±
1
0
∘
,
0
∘
In step 2 , subscript k denotes the present value, and subscript ( k − 1 ) denotes the previous value. E ^ is a unit-length version of the net electric field at point ( x , y , z ) . The parameter ϵ is a very small number.
What is the combined length of all 2 5 field lines?
Details and Assumptions:
1)
4
π
ϵ
0
1
=
1
2)
To check results, it helps to run the plotter with
ϵ
values spanning several orders of magnitude.
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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.
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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.
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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?
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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.
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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.