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Suppose the electric potential due to an electric field is given as $x^2-y^2$, then what will be the graph of the electric field?

My attempt:

Differentiating the potential partially with $x$ and then with $y$ and multiplying with $-1$ I get the electric field expression $E=-2x i +2y j$. After this I am not able to plot the graph.

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2 Answers 2

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One way you can plot the graph is by finding the locus of points on an equipotential surface and then plotting it's orthogonal trajectory.

Let the potential be constant, $x^2-y^2=c$. This represents a hyperbola, find its derivative and equate the derivative of the electric field to the negative reciprocal of the derivative of the hyperbola you just found( as the equipotential surface are perpendicular to the electric field). This will give you the locus of the electric field.

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If you plot some equipotential lines with $x^2-y^2$ equal to $0,\pm 0.5,\pm 1, \pm 2$ etc then the field lines you can draw as being perpendicular to the equipotential lines.

Note the magnitude of the field will vary along a field line.

enter image description here

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