Irodov 3.52: Space Charge That Silences a Plate – Amit & Deepali Sinha Foundation

Problem Irodov 3.52

The gap of width $d$ between two large parallel plates is filled with a uniform volume charge, and the potential difference between the plates is $\Delta\varphi$ . For what charge density $\rho$ does the electric field vanish next to one of the plates? What is the field next to the other plate in that case?

Two large parallel plates, edge on, a distance d apart, the gap filled with a uniform space charge ρ (drawn positive). — Two large parallel plates, edge on, a distance $d$ apart, the gap filled with a uniform space charge $\rho$ (drawn positive).

Given

$d$
$\Delta\varphi$ between the plates
$\rho$ uniform in the gap

Find

\rho

for

E = 0

at one plate;

E

at the other plate

Solution

Step 1The field across the gap

Measure $x$ across the gap from one plate, $0 \le x \le d$ , with $E$ the field component along $x$ . Gauss’s law in differential form, $dE/dx = \rho/\varepsilon_0$ , has a constant right side, so $E$ is linear in $x$ :

E(x) = E(0) + \frac{\rho x}{\varepsilon_0}

Choose the plate at $x = 0$ to be the one where the field vanishes. Then $E(0) = 0$ and

E(x) = \frac{\rho x}{\varepsilon_0} \tag{1}

Step 2The potential difference fixes the charge

The potential drops along the field, so the difference between the plates is the integral of (1) across the gap:

\Delta\varphi = \varphi(0) - \varphi(d) = \int_0^d \frac{\rho x}{\varepsilon_0}\,dx = \frac{\rho d^2}{2\varepsilon_0}

Solving for the density:

\rho = \frac{2\varepsilon_0\,\Delta\varphi}{d^2}

(2)

For $\rho > 0$ the field vanishes at the plate of higher potential; a negative charge makes it vanish at the lower one.

Step 3The field at the other plate

At $x = d$ , (1) gives $\rho d/\varepsilon_0$ , and putting in the density (2):

E(d) = \frac{d}{\varepsilon_0}\cdot\frac{2\varepsilon_0\,\Delta\varphi}{d^2}

E = \frac{\rho d}{\varepsilon_0} = \frac{2\,\Delta\varphi}{d}

(3)

Across the gap, for ρ > 0, with x from the zero-field plate: E (crimson) rises linearly from zero, and φ (teal) leaves that plate flat and falls fastest at the far plate x = d. — Across the gap, for $\rho > 0$ , with $x$ from the zero-field plate: $E$ (crimson) rises linearly from zero, and $\varphi$ (teal) leaves that plate flat and falls fastest at the far plate $x = d$ .

Check

Poisson’s equation gives both answers from the potential alone. The parabola that drops by $\Delta\varphi$ across the gap and is flat at $x = 0$ (zero field there) is

\varphi(x) = \Delta\varphi\left(1 - \frac{x^2}{d^2}\right)

Its second derivative is $-2\Delta\varphi/d^2$ , which equals $-\rho/\varepsilon_0$ for exactly the density (2), and its slope at $x = d$ is $-2\Delta\varphi/d$ , the field (3). Units: $\varepsilon_0\Delta\varphi/d^2$ is $(\text{C}/\text{V m})(\text{V}/\text{m}^2) = \text{C}/\text{m}^3$ .

Remark

A plate next to which the field vanishes carries no charge on its inner face, since that charge is $\varepsilon_0$ times the field at the face. All the field lines from the space charge then end on the other plate, whose inner face carries $-\rho d$ per unit area, the gap’s charge exactly mirrored. The same condition, zero field at an electrode, sets the space-charge-limited current of a vacuum diode; there the charge is not uniform, but the step from Poisson’s equation to the boundary is the same.