Irodov 3.50: Three Components, One Hidden Saddle – Amit & Deepali Sinha Foundation

Problem Irodov 3.50

An electrostatic field is $\vect{E} = ay\vect{i} + (ax + bz)\vect{j} + by\vect{k}$ , where $a$ and $b$ are constants and $\vect{i}$ , $\vect{j}$ , $\vect{k}$ are the unit vectors along the $x$ , $y$ and $z$ axes. Find the potential $\varphi(x, y, z)$ of this field.

Given

$\vect{E} = ay\vect{i} + (ax + bz)\vect{j} + by\vect{k}$
$a$ , $b$ constants

Find

the potential

\varphi(x, y, z)

Solution

Step 1Three equations for the potential

Since $\vect E = -\nabla\varphi$ , the three components of the field give the three partial derivatives of $\varphi$ :

\begin{gathered} \frac{\partial\varphi}{\partial x} = -ay\\ \frac{\partial\varphi}{\partial y} = -(ax + bz)\\ \frac{\partial\varphi}{\partial z} = -by \end{gathered} \tag{1}

Step 2Integrating along x

Integrate the first equation of (1) over $x$ with $y$ and $z$ held fixed. The constant of integration is an unknown function of the other two coordinates:

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

Step 3Fixing the dependence on y

Differentiate (2) with respect to $y$ and match the second equation of (1). The $ax$ terms cancel, and integrating the remainder over $y$ leaves a function of $z$ alone:

\begin{aligned} -ax + \frac{\partial f}{\partial y} &= -ax - bz\\ \frac{\partial f}{\partial y} &= -bz\\ f(y, z) &= -byz + g(z) \end{aligned}

so that

\varphi = -axy - byz + g(z) \tag{3}

Step 4Fixing the dependence on z

Differentiate (3) with respect to $z$ and match the third equation of (1):

\begin{aligned} -by + g'(z) &= -by\\ g'(z) &= 0 \end{aligned}

So $g$ is a constant, and collecting the factor $y$ in (3):

\varphi = -y\,(ax + bz) + \text{const}

(4)

Step 5The shape of the field

With $c = \sqrt{a^2 + b^2}$ , introduce the unit vector $\vect u = (a\vect i + b\vect k)/c$ in the $x$ , $z$ plane and the coordinate $u$ along it. Then the combination is $c$ times $u$ and the answer becomes

\begin{aligned} ax + bz &= c\,u\\ \varphi &= -c\,uy + \text{const} \end{aligned}

This is the potential $-axy$ of a planar saddle field in the $u$ , $y$ plane, with $a$ replaced by $c$ , and nothing changes along the direction perpendicular to both $\vect u$ and $\vect j$ .

The field in the plane of the u and y axes, for any a and b not both zero, and identically in every parallel plane: equipotentials uy = const (teal) and field lines (crimson) crossing them at right angles. — The field in the plane of the $u$ and $y$ axes, for any $a$ and $b$ not both zero, and identically in every parallel plane: equipotentials $uy = \text{const}$ (teal) and field lines (crimson) crossing them at right angles.

Check

Differentiating the answer back gives $-\partial\varphi/\partial x = ay$ , $-\partial\varphi/\partial y = ax + bz$ and $-\partial\varphi/\partial z = by$ , the three given components. The field is curl free, as it must be for a potential to exist: $\partial E_y/\partial x = a = \partial E_x/\partial y$ , $\partial E_z/\partial y = b = \partial E_y/\partial z$ , and $\partial E_x/\partial z = 0 = \partial E_z/\partial x$ . With $b = 0$ the answer reduces to $-axy$ , the potential of the field $a(y\vect i + x\vect j)$ .

Remark

All second derivatives $\partial^2\varphi/\partial x^2$ , $\partial^2\varphi/\partial y^2$ , $\partial^2\varphi/\partial z^2$ vanish, so Laplace’s equation holds and the region is free of charge. Any linear field in a charge-free region reduces this way: its potential is a quadratic form, and turning the axes to the form’s principal directions shows it as $(\lambda_1\xi^2 + \lambda_2\eta^2 + \lambda_3\zeta^2)/2$ with the three $\lambda$ summing to zero. Here they are $c$ , $-c$ and $0$ : a planar saddle, uniform along the third direction.