A problem in graphing electric field lines

  • #1
patric44
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Homework Statement:
a problem in Graphing electric field lines
Relevant Equations:
dy/dx = Ey/Ex
hi guys
our instructor asked us to try to graph the projection of the electric field intensity at a certain point p(x,y) , for two charges q+-q located
at (-a,0) , (a,0), Now starting with the equation
$$\frac{dy}{dx} = \frac{E_{y}}{E_{x}}$$
after transforming this equation I got
$$\frac{du}{(1+u^{2})^{3/2}}=\frac{dv}{(1+v^{2})^{3/2}}$$

where $u = (x+a)/y$ , $v = (x-a)/y$

my question is how would I plot the solution of this equation to get the graph of the electric field?
 

Answers and Replies

  • #2
kuruman
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Do you know how to find orthogonal trajectories? If yes, find the orthogonal trajectories to the equipotential lines of the two charges. If no, look up how to do it on the internet. There are many references. Here is one with examples.
 
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  • #3
patric44
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so basically i have to solve this equation to get the electric field ?
$$ \frac{du}{dv} = \frac{(v^2+1)^{3/2}}{(u^2+1)^{3/2}}$$
another question : how to plot these family of curves say in Matlab
 
  • #4
anuttarasammyak
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Relevant Equations:: dy/dx = Ey/Ex

after transforming this equation I got
You may integrate the both sides of this equation to get u, v relation. Substitution u, v = tan ##\theta## seems to work.
 
  • #5
bob012345
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Homework Statement:: a problem in Graphing electric field lines
Relevant Equations:: dy/dx = Ey/Ex

hi guys
our instructor asked us to try to graph the projection of the electric field intensity at a certain point p(x,y) , for two charges q+-q located
at (-a,0) , (a,0), Now starting with the equation
$$\frac{dy}{dx} = \frac{E_{y}}{E_{x}}$$
after transforming this equation I got
$$\frac{du}{(1+u^{2})^{3/2}}=\frac{dv}{(1+v^{2})^{3/2}}$$

where $u = (x+a)/y$ , $v = (x-a)/y$

my question is how would I plot the solution of this equation to get the graph of the electric field?
Please define what you mean by the projection of the electric field intensity at a certain point p(x,y) because the actual electric field of two point charges is easy to write down. And what does it have to do with ##\frac{dy}{dx} = \frac{E_{y}}{E_{x}}##?
 
  • #6
kuruman
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so basically i have to solve this equation to get the electric field ?
$$ \frac{du}{dv} = \frac{(v^2+1)^{3/2}}{(u^2+1)^{3/2}}$$
another question : how to plot these family of curves say in Matlab
How are ##u## and ##v## related to ##x##, ##y## and ##a##?
 
  • #7
patric44
272
37
You may integrate the both sides of this equation to get u, v relation. Substitution u, v = tan ##\theta## seems to work.
My confusion is how to plot the family of curves after integrating both sides.
 
  • #8
patric44
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37
Please define what you mean by the projection of the electric field intensity at a certain point p(x,y) because the actual electric field of two point charges is easy to write down. And what does it have to do with ##\frac{dy}{dx} = \frac{E_{y}}{E_{x}}##?
Image 1.jpg

I wish to get something similar to this graph after solving
$$\frac{dy}{dx}=\frac{ \frac{kq\left(x-a\right)}{\left[\left(x-a\right)^2+y^2\right]^{3/2}}\pm \frac{kq\left(x+a\right)}{\left[\left(x+a\right)^2+y^2\right]^{3/2}}}{\frac{kqy}{\left[\left(x-a\right)^2+y^2\right]^{3/2}}\pm\frac{kqy}{\left[\left(x+a\right)^2+y^2\right]^{3/2}}}$$
this is the equation before the transformation
$$u=\frac{x+a}{y}, v=\frac{x-a}{y}.$$
 
  • #9
patric44
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How are ##u## and ##v## related to ##x##, ##y## and ##a##?
Through the transformation
$$u=\frac{x+a}{y}, v=\frac{x-a}{y}$$
 
  • #10
anuttarasammyak
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My confusion is how to plot the family of curves after integrating both sides.
Thus got u-v relation is rewritten as x-y relation , I expect.
 
  • #11
patric44
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Thus got u-v relation is rewritten as x-y relation , I expect.
Yes I tried that, but it doesn't look like the graph shown above! Also, I am trying to figure some way to graph an implicit family of curves using MATLAB or Python.
 
  • #12
anuttarasammyak
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Could you show us your integral result of u, v functions ? That seems not so much complicated for writing down.
 
Last edited:
  • #13
bob012345
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It seems to me you are not solving for the electric field, you are trying to plot the field lines and lines of equal potential. Is that correct?
 
  • #14
bob012345
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View attachment 290951
I wish to get something similar to this graph after solving
$$\frac{dy}{dx}=\frac{ \frac{kq\left(x-a\right)}{\left[\left(x-a\right)^2+y^2\right]^{3/2}}\pm \frac{kq\left(x+a\right)}{\left[\left(x+a\right)^2+y^2\right]^{3/2}}}{\frac{kqy}{\left[\left(x-a\right)^2+y^2\right]^{3/2}}\pm\frac{kqy}{\left[\left(x+a\right)^2+y^2\right]^{3/2}}}$$
this is the equation before the transformation
$$u=\frac{x+a}{y}, v=\frac{x-a}{y}.$$
I still don't see where you got this formula ##\frac{dy}{dx}= \frac{E_y}{E_x}## and what it represents. You already know the electric field everywhere so aren't you merely trying to plot the field knowing the strength at each point? It seems to me there are an infinite number of lines you could plot so how are you going to select which ones?
 
Last edited:
  • #15
anuttarasammyak
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I still don't see where you got this formula dydx=EyEx and what it represents.
Say line element (dX,dY) is perpendicular to E-field their inner product is zero.
[tex] E_xdX+E_ydY=0[/tex]
[tex]dY/dX=-E_x/E_y[/tex]
The lines perpendicular to (dX, dY), so tangent to E everywhere on the line, has tangent of minus reciprocal of the above
[tex]dy/dx=E_y/E_x[/tex]
 
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  • #17
anuttarasammyak
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Substitution u, v = tan θ seems to work.
[tex]\int \frac{du}{(1+u^2)^{3/2}}= \int \frac{\sec^2\theta\ d\theta}{\sec^3\theta} [/tex]
 
Last edited:
  • #18
bob012345
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[tex]\int \frac{du}{(1+u^2)^{3/2}}= \int \frac{\sec^2\theta\ d\theta}{\sec^3\theta} [/tex]
I think one can plot families of curves for like and unlike charges by assuming values for ##\frac{dy}{dx}##.
desmos-graph (25).png


desmos-graph (25).png
 

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