Kvant Physics Problem 170

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Verdicts: PASS + PASS
Solve time: 2m26s
Source on kvant.digital

Problem

In a homogeneous plasma with density (the number of charges of each sign per unit volume — in $1~\text{см}^3$) $n$ all electrons, initially located in a layer of thickness $x$, are displaced along the normal to this layer by a distance $x$. Find the electric field $E$ in the cross section $SS$ (Fig. 2).

Figure 2

V. G. Averin

Setup and Assumptions

A homogeneous plasma initially occupies a slab of thickness $x$ and contains a uniform density $n$ of positive ions and an equal density $n$ of electrons, where $n$ is the number of particles of each sign per unit volume, given in $\text{cm}^{-3}$. The plasma is electrically neutral in its initial state.

All electrons in the slab are displaced rigidly along the normal to the slab by a distance $x$, while the positive ions remain fixed in space. After the displacement, the electrons form a slab parallel to the ion slab. Each slab has the same thickness $x$ and uniform volume charge density magnitude $en$, where $e$ is the elementary charge.

The field is required in the cross section $SS$, located in the region between the displaced electron slab and the ion slab. Edge effects are neglected, so both slabs are treated as infinite parallel layers.

Physical Principles

The electric field of a uniformly charged infinite slab follows from Gauss’s law. In Gaussian units, the normal component of the electric field satisfies

$\oint \mathbf{E}\cdot d\mathbf{S} = 4\pi Q_{\text{enc}}.$

For an infinite plane sheet with surface charge density $\sigma$, the field magnitude is

$E = 2\pi \sigma$

on each side of the sheet, directed away from positive charge and toward negative charge.

For multiple parallel charged sheets, the net field is obtained by linear superposition of the individual fields.

The surface charge density of a uniformly filled slab of thickness $x$ and volume charge density $\rho$ is

$\sigma = \rho x.$

Derivation

After the displacement, the ion slab carries a uniform positive charge density $\rho_+ = en$, and the electron slab carries a uniform negative charge density $\rho_- = -en$. Each slab has thickness $x$, so the corresponding surface charge densities at their facing surfaces are

$\sigma = enx.$

The ion slab produces an electric field of magnitude $2\pi enx$, directed away from the ion slab. The electron slab produces an electric field of the same magnitude $2\pi enx$, directed toward the electron slab.

In the region between the slabs, both contributions are directed from the ion slab toward the electron slab, so the fields add linearly. The total electric field in the cross section $SS$ is therefore

$E = 2\pi enx + 2\pi enx = 4\pi enx.$

The direction of the field is along the normal from the positively charged ion layer toward the negatively charged electron layer.

Result

The electric field in the cross section $SS$ is

$E = 4\pi e n x.$

In Gaussian units, $e$ is measured in statcoulombs and $n$ in $\text{cm}^{-3}$, giving $E$ in $\text{statV/cm}$. Substituting a symbolic density $n = n,\text{cm}^{-3}$ and displacement $x = x,\text{cm}$ yields

$E = 4\pi e n x ;; \text{statV/cm}.$

Sanity Checks

The expression is linear in $n$, so doubling the plasma density doubles the electric field, consistent with Gauss’s law for superposed charge distributions. The field is also linear in $x$, since a thicker initial layer produces a larger surface charge density $enx$ after separation.

Dimensional consistency follows from $\sigma = enx$, which has units of charge per area, and the Gaussian relation $E \sim \sigma$, giving the correct electric field units.

If $x \to 0$, the charge separation vanishes and $E \to 0$, matching the neutral plasma limit. If $n \to 0$, no charges are present and the field also vanishes. A sign error would most likely arise from assigning both slabs fields in opposite directions in the inter-slab region, which would incorrectly predict cancellation instead of reinforcement.