Irodov 3.51: Charge from the Curvature of Potential – Amit & Deepali Sinha Foundation

Problem Irodov 3.51

In some region of space the electrostatic potential depends on the coordinate $x$ alone, as $\varphi = -ax^3 + b$ with $a$ and $b$ constants. Find the volume charge density $\rho(x)$ in this region.

Given

$\varphi = -ax^3 + b$
$a$ , $b$ constants

Find

the space charge density

\rho(x)

Solution

Step 1The field from the potential

The potential depends on $x$ alone, so the field has only an $x$ component, minus the slope of $\varphi$ :

E_x = -\frac{d\varphi}{dx} = -\frac{d}{dx}\left(-ax^3 + b\right) = 3ax^2 \tag{1}

The constant $b$ has dropped out: it shifts the potential and changes nothing physical.

For a > 0, with b = 0: the field E_x (crimson) is positive on both sides of O, smallest at O, and the potential φ (teal) falls throughout, steepest far from O. — For $a > 0$ , with $b = 0$ : the field $E_x$ (crimson) is positive on both sides of $O$ , smallest at $O$ , and the potential $\varphi$ (teal) falls throughout, steepest far from $O$ .

Step 2The charge from the field

In differential form Gauss’s law is $\nabla\cdot\vect E = \rho/\varepsilon_0$ , and with $E_x$ the only component it reduces to one derivative. Using (1):

\rho = \varepsilon_0\,\frac{dE_x}{dx} = \varepsilon_0\,\frac{d}{dx}\left(3ax^2\right)

\rho = 6\varepsilon_0 a x

(2)

Check

Poisson’s equation gives the same result without the field: $d^2\varphi/dx^2 = -6ax$ , and $\rho = -\varepsilon_0\,d^2\varphi/dx^2 = 6\varepsilon_0 ax$ . Units: $a$ is volts per cubic metre, since $ax^3$ is a potential, so $\varepsilon_0 a x$ has the units $(\text{C}/\text{V m})(\text{V}/\text{m}^3)\,\text{m}$ , which is $\text{C}/\text{m}^3$ , a charge density.

Remark

The charge is odd in $x$ : for $a > 0$ it is negative at $x < 0$ and positive at $x > 0$ , so the region as a whole is a double layer. The field stays positive on both sides of $O$ because it is fed by the charge already crossed: the negative layer brings it down to zero at $O$ , and the positive layer builds it back up, which is why the graph of $E_x$ has its minimum exactly where $\rho$ changes sign.