Irodov 3.48: Potential Behind a Saddle Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.48

An electrostatic field is $\vect{E} = a(y\vect{i} + x\vect{j})$ , where $a$ is a constant and $\vect{i}$ , $\vect{j}$ are the unit vectors along the $x$ and $y$ axes. Find the potential $\varphi(x, y)$ of this field.

Given

$\vect{E} = a(y\vect{i} + x\vect{j})$
$a$ constant

Find

the potential

\varphi(x, y)

Solution

Step 1The two equations for the potential

The field is minus the gradient of the potential, so each component of $\vect E$ fixes one partial derivative of $\varphi$ . With $E_x = ay$ and $E_y = ax$ :

\begin{gathered} \frac{\partial\varphi}{\partial x} = -E_x = -ay\\ \frac{\partial\varphi}{\partial y} = -E_y = -ax \end{gathered} \tag{1}

The field has no $z$ component, so $\varphi$ does not depend on $z$ .

Step 2Integrating along x

Integrate the first equation of (1) over $x$ at fixed $y$ . The constant of integration may still depend on $y$ , so it is an unknown function $f(y)$ :

\varphi = -axy + f(y) \tag{2}

Step 3Fixing the function of y

Differentiate (2) with respect to $y$ and set the result equal to the second equation of (1):

\begin{aligned} \frac{\partial\varphi}{\partial y} = -ax + f'(y) &= -ax\\ f'(y) &= 0 \end{aligned}

So $f$ is a pure constant, and (2) becomes

\varphi = -axy + \text{const}

(3)

Step 4The shape of the field

The equipotentials are the rectangular hyperbolas $xy = \text{const}$ , with the axes as the level through the origin. A field line runs along $\vect E$ , so its slope is the ratio of the components, and the variables separate:

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

y^2 - x^2 = C

(4)

For $a > 0$ the field points up across the positive $x$ axis, where $\vect E = ax\,\vect j$ , and right across the positive $y$ axis, where $\vect E = ay\,\vect i$ .

The field for a > 0: equipotentials xy = const (teal) and field lines y^2 - x^2 = C (crimson), crossing at right angles. The field vanishes at the origin, where the diagonal lines cross. — The field for $a > 0$ : equipotentials $xy = \text{const}$ (teal) and field lines $y^2 - x^2 = C$ (crimson), crossing at right angles. The field vanishes at the origin, where the diagonal lines cross.

Check

Differentiating the answer back, $-\partial\varphi/\partial x = ay = E_x$ and $-\partial\varphi/\partial y = ax = E_y$ . That a potential exists at all needs a curl-free field, $\partial E_y/\partial x = \partial E_x/\partial y$ , and here both sides equal $a$ . A field such as $a(y\vect i - x\vect j)$ would give $-a$ against $a$ , and no $\varphi$ could reproduce it.

Remark

The potential is fixed only up to an additive constant, which is why the answer carries one: a field measures differences of potential, never its value. This potential also satisfies Laplace’s equation, $\partial^2\varphi/\partial x^2 + \partial^2\varphi/\partial y^2 = 0 + 0$ , so the region holds no charge, and the origin, where $\vect E = 0$ , is a saddle of $\varphi$ : in a charge-free region the potential can have no maximum or minimum.