Irodov 3.49: Cubic Potential, Built Term by Term – Amit & Deepali Sinha Foundation

Problem Irodov 3.49

An electrostatic field is $\vect{E} = 2axy\vect{i} + a(x^2 - y^2)\vect{j}$ , 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

$\vect{E} = 2axy\vect{i} + a(x^2 - y^2)\vect{j}$
$a$ constant

Find

the potential

\varphi(x, y)

Solution

Step 1The two equations for the potential

Since $\vect E = -\nabla\varphi$ , each component of the field gives one partial derivative of $\varphi$ ; the field has no $z$ component, so $\varphi$ depends on $x$ and $y$ only.

\begin{gathered} \frac{\partial\varphi}{\partial x} = -2axy\\ \frac{\partial\varphi}{\partial y} = -a(x^2 - y^2) \end{gathered} \tag{1}

Step 2Integrating along x

Integrate the first equation of (1) over $x$ at fixed $y$ ; the constant of integration is an unknown function $f(y)$ :

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

Step 3Fixing the function of y

Differentiate (2) with respect to $y$ and match it to the second equation of (1). The $x^2$ terms agree, and what is left fixes $f'$ :

\begin{aligned} -ax^2 + f'(y) &= -ax^2 + ay^2\\ f'(y) &= ay^2\\ f(y) &= \frac{ay^3}{3} + \text{const} \end{aligned}

Putting $f$ back into (2):

\varphi = ay\left(\frac{y^2}{3} - x^2\right) + \text{const}

(3)

Step 4The potential along the y axis

On the line $x = 0$ the answer and the field reduce to one variable each:

\begin{gathered} \varphi(0, y) = \frac{ay^3}{3} + \text{const}\\ E_y(0, y) = -ay^2 \end{gathered}

For $a > 0$ the potential rises along the whole axis, so the field points down the axis on both sides of the origin $O$ , strongest far out and vanishing at $O$ , where the potential has a flat inflection.

Along the y axis, for a > 0 with the constant set to zero: E_y (crimson) is negative on both sides of O, and φ (teal) rises throughout, flattening at O. — Along the $y$ axis, for $a > 0$ with the constant set to zero: $E_y$ (crimson) is negative on both sides of $O$ , and $\varphi$ (teal) rises throughout, flattening at $O$ .

Check

Differentiating the answer back, $-\partial\varphi/\partial x = 2axy = E_x$ and $-\partial\varphi/\partial y = -ay^2 + ax^2 = E_y$ , the given field. The field is curl free, as an electrostatic field must be: $\partial E_y/\partial x = 2ax$ and $\partial E_x/\partial y = 2ax$ . On the graph, $E_y$ is minus the slope of $\varphi$ at every $y$ .

Remark

The second derivatives, $\partial^2\varphi/\partial x^2 = -2ay$ and $\partial^2\varphi/\partial y^2 = 2ay$ , cancel, so this potential obeys Laplace’s equation and the region is free of charge. Its zero level is three straight lines through $O$ , $y = 0$ and $y = \pm\sqrt{3}\,x$ , sixty degrees apart: $O$ is a three-fold (monkey) saddle, the next shape after the ordinary saddle of a quadratic potential.