# A problem in graphing electric field lines

• patric44
In summary: The equation you are trying to solve is$$\frac{du}{dv}=\frac{(v^2+1)^{3/2}}{(u^2+1)^{3/2}}and it represents the electric field intensity at a point. You are trying to graph the field by solving for the intensity at a point. #### patric44 Homework Statement a problem in Graphing electric field lines Relevant Equations dy/dx = Ey/Ex hi guys our instructor asked us to try to graph the projection of the electric field intensity at a certain point p(x,y) , for two charges q+-q located at (-a,0) , (a,0), Now starting with the equation$$\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}}$$where u = (x+a)/y , v = (x-a)/y my question is how would I plot the solution of this equation to get the graph of the electric field? Do you know how to find orthogonal trajectories? If yes, find the orthogonal trajectories to the equipotential lines of the two charges. If no, look up how to do it on the internet. There are many references. Here is one with examples. berkeman and patric44 so basically i have to solve this equation to get the electric field ?$$ \frac{du}{dv} = \frac{(v^2+1)^{3/2}}{(u^2+1)^{3/2}}$$another question : how to plot these family of curves say in Matlab patric44 said: Relevant Equations:: dy/dx = Ey/Ex after transforming this equation I got You may integrate the both sides of this equation to get u, v relation. Substitution u, v = tan ##\theta## seems to work. patric44 said: Homework Statement:: a problem in Graphing electric field lines Relevant Equations:: dy/dx = Ey/Ex hi guys our instructor asked us to try to graph the projection of the electric field intensity at a certain point p(x,y) , for two charges q+-q located at (-a,0) , (a,0), Now starting with the equation$$\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}}$$where u = (x+a)/y , v = (x-a)/y my question is how would I plot the solution of this equation to get the graph of the electric field? Please define what you mean by the projection of the electric field intensity at a certain point p(x,y) because the actual electric field of two point charges is easy to write down. And what does it have to do with ##\frac{dy}{dx} = \frac{E_{y}}{E_{x}}##? patric44 said: so basically i have to solve this equation to get the electric field ?$$ \frac{du}{dv} = \frac{(v^2+1)^{3/2}}{(u^2+1)^{3/2}}$$another question : how to plot these family of curves say in Matlab How are ##u## and ##v## related to ##x##, ##y## and ##a##? anuttarasammyak said: You may integrate the both sides of this equation to get u, v relation. Substitution u, v = tan ##\theta## seems to work. My confusion is how to plot the family of curves after integrating both sides. bob012345 said: Please define what you mean by the projection of the electric field intensity at a certain point p(x,y) because the actual electric field of two point charges is easy to write down. And what does it have to do with ##\frac{dy}{dx} = \frac{E_{y}}{E_{x}}##? I wish to get something similar to this graph after solving$$\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}.$$kuruman said: How are ##u## and ##v## related to ##x##, ##y## and ##a##? Through the transformation$$u=\frac{x+a}{y}, v=\frac{x-a}{y}$$patric44 said: My confusion is how to plot the family of curves after integrating both sides. Thus got u-v relation is rewritten as x-y relation , I expect. anuttarasammyak said: Thus got u-v relation is rewritten as x-y relation , I expect. Yes I tried that, but it doesn't look like the graph shown above! Also, I am trying to figure some way to graph an implicit family of curves using MATLAB or Python. Could you show us your integral result of u, v functions ? That seems not so much complicated for writing down. Last edited: It seems to me you are not solving for the electric field, you are trying to plot the field lines and lines of equal potential. Is that correct? patric44 said: View attachment 290951 I wish to get something similar to this graph after solving$$\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?

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bob012345 said:
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.
$$E_xdX+E_ydY=0$$
$$dY/dX=-E_x/E_y$$
The lines perpendicular to (dX, dY), so tangent to E everywhere on the line, has tangent of minus reciprocal of the above
$$dy/dx=E_y/E_x$$

Delta2, PhDeezNutz and bob012345
anuttarasammyak said:
Substitution u, v = tan θ seems to work.
$$\int \frac{du}{(1+u^2)^{3/2}}= \int \frac{\sec^2\theta\ d\theta}{\sec^3\theta}$$

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anuttarasammyak said:
$$\int \frac{du}{(1+u^2)^{3/2}}= \int \frac{\sec^2\theta\ d\theta}{\sec^3\theta}$$
I think one can plot families of curves for like and unlike charges by assuming values for ##\frac{dy}{dx}##.

PhDeezNutz

## 1. What is a problem in graphing electric field lines?

A problem in graphing electric field lines is when there is difficulty in accurately representing the direction and strength of the electric field at different points on a graph.

## 2. Why is it important to accurately graph electric field lines?

Accurately graphing electric field lines is important because it helps us visualize and understand the behavior of electric fields, which are crucial in many areas of science and technology.

## 3. What factors can affect the accuracy of graphing electric field lines?

Factors that can affect the accuracy of graphing electric field lines include the distance between charges, the magnitude of the charges, and the presence of other charges or objects in the surrounding area.

## 4. How can we improve the accuracy of graphing electric field lines?

To improve the accuracy of graphing electric field lines, we can use more precise measuring tools, increase the number of data points on the graph, and consider the effects of other charges or objects in the area.

## 5. What are some common misconceptions about graphing electric field lines?

Some common misconceptions about graphing electric field lines include thinking that the lines represent the path of a charged particle, or that the lines themselves have physical properties. In reality, electric field lines are simply a visual representation of the electric field at different points in space.