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I've seen that if you wanted to plot the flow lines (field lines) of a vector field $F$, then a simple method is to find expressions for the x and y components of the field: $E_x$ and $E_y$ respectively, and then form an equation:

$$\frac{dy}{dx}=\frac{E_y}{E_x}$$

Then integrate (solve the ODE). This would then give you the general form of the flow lines (a set of equations where you can vary the constant of integration to get all flow lines).

My question is, given an electric field formed by two arbitrary point charges in space, if you wished to plot the flow lines of the field, could you find $\frac{E_y}{E_{x}}$ for charge 1 and then find the same expression for charge two and then add them together to find $\frac{dy}{dx}$, according to superposition or would this not give the correct expression? If it wouldn't give the right flow lines, why would it not?

Mathematically, if we use superscript notation for components due to charges 1 and 2.

Combining the components: $$\frac{dy}{dx}=\frac{E_y^1+E_y^2}{E_x^1+E_x^2}$$ Which is definitely correct.

Adding the gradients of the fields individually: $$\frac{dy}{dx}=\frac{E_y^1}{E_x^1}+\frac{E_y^2}{E_x^2}$$ Is this equivalent? It seems it should not be just by considering general fractions, but for the specific case of electric fields due to separate charges, is it equivalent due to linearity?

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You have to be careful where the linearity comes from. The linearity comes from the Maxwell equation in the vacuum. Therefore you can just add the two solutions of the equations around the point charges and get a valid electric field for all the vacuum around.

Your transformation $E_x / E_y$ is not a linear transformation. So although the Maxwell equations are linear, you lose the linearity there.

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  • $\begingroup$ Ah I see - thanks! Expressing it in terms of the transformation makes things a lot clearer. Also, my original reason for thinking this was the mistaken idea that adding the fields seemed to add the directions of the field components, but now I realise that it actually has to add them in component form and not overall as $\frac{E_y}{E_x}$, as suggested by the maths. $\endgroup$ – Resquiens Feb 12 '17 at 17:57

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