Kvant Physics Problem 170

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Problem

In a homogeneous plasma with density (number of charges of each sign per unit volume — in $1~\text{см}^3$) $n$, all electrons that were initially located in a layer of thickness $x$ are displaced 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 contains electrons and ions with equal number density $n$ in $\text{cm}^{-3}$. The ions are assumed fixed in space, while the electrons are displaced rigidly along a direction normal to a chosen cross-section $SS$. A layer of plasma of thickness $x$ is considered, and all electrons originally contained in this layer are shifted by a distance $x$ along the same normal direction.

The plasma is assumed infinite in the transverse directions so that all fields depend only on the coordinate along the displacement direction. Edge effects at the boundaries of the layer are neglected. The system is treated in Gaussian units, consistent with the use of density in $\text{cm}^{-3}$. The goal is to determine the electric field $E$ at the cross-section $SS$, which lies in the region of charge separation created by the displacement.

Physical Principles

Gauss’s law in differential form relates the divergence of the electric field to the charge density,

$$\nabla \cdot \mathbf{E} = 4\pi \rho.$$

For planar symmetry, the electric field is directed along a single coordinate axis, and its magnitude depends only on position along that axis. In this case Gauss’s law reduces to the relation between electric field discontinuity and surface charge density,

$$E_{\perp} = 2\pi \sigma$$

for a single infinite charged sheet in Gaussian units.

Superposition applies to electric fields produced by multiple charge distributions. The charge density generated by separating electrons and ions can be represented as equivalent charged slabs or sheets with uniform density derived from the displacement of charge carriers.

Derivation

The plasma initially has equal ion and electron number densities $n$, so the initial charge density is zero everywhere.

Consider a layer of thickness $x$. After displacement, all electrons originally occupying this layer are shifted by distance $x$ along the normal direction. The ions remain fixed. This produces a region of thickness $x$ from which electrons are removed, leaving an uncompensated positive charge density

$$\rho_+ = +en,$$

where $e$ is the elementary charge.

At the destination region of the displaced electrons, an additional electron density appears equal to $n$, producing a negative charge density

$$\rho_- = -en$$

in that region over the same thickness $x$.

Each of these regions can be replaced by an equivalent infinite slab of uniform charge density. A slab of thickness $x$ with uniform volume charge density $\rho$ is equivalent, as seen from outside the slab, to a surface charge density

$$\sigma = \rho x.$$

Thus each region produces an effective surface charge magnitude

$$\sigma = enx.$$

The configuration consists of two parallel charged sheets of equal magnitude and opposite sign separated by a distance $x$. The electric field between two infinite parallel sheets in Gaussian units equals the sum of the fields from each sheet. Each sheet produces field magnitude $2\pi\sigma$, and between opposite sheets the fields add, giving

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

Substituting $\sigma = enx$ gives

$$E = 4\pi enx.$$

The direction of the field is from the positively charged region toward the negatively charged region.

Result

The electric field in the cross-section $SS$ is

$$E = 4\pi e n x.$$

Substituting $n$ in $\text{cm}^{-3}$, $x$ in $\text{cm}$, and $e = 4.803 \times 10^{-10}\ \text{statC}$,

$$E = 4\pi \cdot (4.803 \times 10^{-10}\ \text{statC}) \cdot n(\text{cm}^{-3}) \cdot x(\text{cm}).$$

Hence,

$$\boxed{E \approx 6.03 \times 10^{-9}, n x \ \text{statV/cm}}$$

with $n$ in $\text{cm}^{-3}$ and $x$ in $\text{cm}$.

Sanity Checks

Dimensional consistency follows from $enx$, where $e$ carries charge, $n$ carries inverse volume, and $x$ carries length, producing surface charge density with units of charge per area, which in Gaussian units yields electric field units after multiplication by $4\pi$.

If the displacement $x$ tends to zero, the separated charge layers vanish and the field reduces to zero linearly in $x$, consistent with weak perturbation behavior.

If the plasma density $n$ increases, the field increases proportionally because both positive and negative charge excesses scale with the number of displaced carriers.

A sign error most easily arises when assigning the effective charge densities of the depleted and accumulated layers, which must be opposite in sign to produce additive fields between the layers.