Question

The lines of electric force of two opposite charges of the same strength at (-1,0) and (1,0) are the circles through (-1,0) and (1,0). Show that these circles are given by x^2 + (y - c)^2 = 1 + c^2. Show that the equipotential lines (which are orthogonal trajectories of those circles) are the circles given by (x + c*)^2 + y^2 = c*^2 - 1.

Answer 1

Solution:

Setting y=0 gives from $x^{2}+(y-c)^{2}=1+c^{2}$ the equation $x^{2}+c^{2}=1+c^{2}$ ; hence x=-1 and x=1, which verifies that those circles all pass through -1 and 1 , each of them simultaneously through both points. Subtracting $c^{2}$ on both sides of the given equation, we obtain

$x^{2}+y^{2}-2 c y=1, \quad x^{2}+y^{2}-1=2 c y, \quad \frac{x^{2}-1}{y}+y=2 c .$

Emphasize to your class that the ODE for the given curves must always be free of c. Having accomplished this, we can now differentiate. This gives

$\frac{2 x}{y}-\left(\frac{x^{2}-1}{y^{2}}-1\right) y^{\prime}=0$

This is the ODE of the given curves. Replacing $y^{\prime}\ with -1 / \widetilde{y}^{\prime}$ and y with $\tilde{y}$ , we obtain the ODE of the trajectories:

$\frac{2 x}{\widetilde{y}}-\left(\frac{x^{2}-1}{\tilde{y}^{2}}-1\right) /\left(-\widetilde{y}^{\prime}\right)=0$

Multiplying this by $y^{\prime}$ , we get

$\frac{2 x \tilde{y}^{\prime}}{\tilde{y}}+\frac{x^{2}-1}{\tilde{y}^{2}}-1=0 .$

Multiplying this by $\widetilde{y}^{2} / x^{2}$ , we obtain

$\frac{2 \widetilde{y} \tilde{y}^{\prime}}{x}+1-\frac{1}{x^{2}}-\frac{\tilde{y}^{2}}{x^{2}}=\frac{d}{d x}\left(\frac{\tilde{y}^{2}}{x}\right)+1-\frac{1}{x^{2}}=0 .$

By integration,

$\frac{\tilde{y}^{2}}{x}+x+\frac{1}{x}=2 c^{*} . \quad \text { Thus, } \quad \tilde{y}^{2}+x^{2}+1=2 c^{*} x$ .

We see that these are the circles

$\tilde{y}^{2}+\left(x-c^{*}\right)^{2}=c^{*}-1$