Irodov 3.36: Hyperbolic Fields from Saddle Potentials

Problem Irodov 3.36

An electrostatic potential depends only on the coordinates $x$ and $y$ , with $a$ a constant. Find the electric field vector $\vect{E}$ in each case, and sketch the field lines in the $x$ , $y$ plane:

$\varphi = a(x^2 - y^2)$ ;
$\varphi = axy$ .

Equipotentials of φ = a(x^2 - y^2) for a > 0, at equal steps of φ: hyperbolas whose asymptotes are the diagonals. — Equipotentials of $\varphi = a(x^2 - y^2)$ for $a > 0$ , at equal steps of $\varphi$ : hyperbolas whose asymptotes are the diagonals.

Equipotentials of φ = axy: the same family turned through 45°, whose asymptotes are the axes. — Equipotentials of $\varphi = axy$ : the same family turned through $45^\circ$ , whose asymptotes are the axes.

Given

(a) $\varphi = a(x^2 - y^2)$
(b) $\varphi = axy$
$a$ constant

Find

\vect{E}

in each case; the field lines in the

x

y

plane

Solution

Step 1Field from potential

The field is minus the gradient of the potential. Neither potential depends on $z$ , so $E_z = 0$ and only two components remain.

\vect{E} = -\frac{\partial\varphi}{\partial x}\,\vect{i} - \frac{\partial\varphi}{\partial y}\,\vect{j} \tag{1}

A field line runs along $\vect{E}$ at every point, so its slope in the $x$ , $y$ plane is the ratio of the two components.

\frac{dy}{dx} = \frac{E_y}{E_x} \tag{2}

Step 2Case (a): the field

For $\varphi = a(x^2 - y^2)$ the two partial derivatives are

\frac{\partial\varphi}{\partial x} = 2ax, \qquad \frac{\partial\varphi}{\partial y} = -2ay

and (1) gives

\vect{E} = -2a\,(x\,\vect{i} - y\,\vect{j})

(3)

Step 3Case (a): the field lines

Putting $E_x = -2ax$ and $E_y = 2ay$ into (2) separates the variables.

\begin{aligned} \frac{dy}{dx} &= -\frac{y}{x}\\ \frac{dy}{y} &= -\frac{dx}{x}\\ \ln\lvert y\rvert &= -\ln\lvert x\rvert + \text{const} \end{aligned}

xy = C

(4)

The lines are rectangular hyperbolas with the axes as asymptotes ( $C = 0$ gives the axes themselves). For $a > 0$ , on the $x$ axis $E_x = -2ax$ points toward the origin, and on the $y$ axis $E_y = 2ay$ points away from it: the lines come in along $x$ and leave along $y$ .

Field lines xy = C of case (a) for a > 0 (crimson), inward along the x axis and outward along y, crossing the teal equipotentials at right angles. — Field lines $xy = C$ of case (a) for $a > 0$ (crimson), inward along the $x$ axis and outward along $y$ , crossing the teal equipotentials at right angles.

Step 4Case (b): the field

For $\varphi = axy$ the partial derivatives are

\frac{\partial\varphi}{\partial x} = ay, \qquad \frac{\partial\varphi}{\partial y} = ax

and (1) gives

\vect{E} = -a\,(y\,\vect{i} + x\,\vect{j})

(5)

Step 5Case (b): the field lines

Now $E_x = -ay$ and $E_y = -ax$ , and (2) again separates.

\begin{aligned} \frac{dy}{dx} &= \frac{x}{y}\\ y\,dy &= x\,dx \end{aligned}

y^2 - x^2 = C

(6)

These are rectangular hyperbolas with the diagonals $y = \pm x$ as asymptotes. For $a > 0$ , on the positive $x$ axis $\vect{E} = -ax\,\vect{j}$ points down and on the negative $x$ axis it points up; on the positive $y$ axis $\vect{E} = -ay\,\vect{i}$ points left and on the negative $y$ axis right.

Field lines y^2 - x^2 = C of case (b) for a > 0: down across the positive x axis, up across the negative one, left across the positive y axis and right across the negative one. — Field lines $y^2 - x^2 = C$ of case (b) for $a > 0$ : down across the positive $x$ axis, up across the negative one, left across the positive $y$ axis and right across the negative one.

Check

An electrostatic field must be curl free, $\partial E_y/\partial x - \partial E_x/\partial y = 0$ . In case (a) both derivatives vanish. In case (b), from (5), $\partial E_y/\partial x = -a$ and $\partial E_x/\partial y = -a$ , and the difference is zero. The sign of the $\vect{j}$ term matters here: a field $-a(y\,\vect{i} - x\,\vect{j})$ would have the difference $2a$ , a rotating field that no potential can produce. As a second check, the field lines of (a), $xy = C$ , cut the equipotentials $x^2 - y^2 = \text{const}$ at right angles, as lines of force must.

Remark

Turning the axes through $45^\circ$ , with $x = (u - v)/\sqrt{2}$ and $y = (u + v)/\sqrt{2}$ , gives $xy = (u^2 - v^2)/2$ , so case (b) is case (a) rotated by $45^\circ$ with $a$ halved. In a charge-free region, a potential independent of $z$ is, near a point where $\vect{E}$ vanishes and to second order, a combination of these two: Laplace’s equation forces its curvatures along $x$ and $y$ to be equal and opposite, so such a point is a saddle, never a minimum (Earnshaw’s theorem in two dimensions). Electrodes shaped along these hyperbolas make the quadrupole lenses and mass filters of particle physics.