Kvant Physics Problem 173

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Verdicts: PASS + PASS
Solve time: 6m30s
Source on kvant.digital

Problem

Published in the journal under number 163 in error.

In a uniformly charged sphere of radius $R$ there is a spherical cavity of radius $r$, whose center is at a distance $a$ from the center of the sphere (Fig. 1). Find the electric field strength at different points inside the cavity if the charge density is $\sigma$.

Figure 2

V. D. Krivchenkov

Setup and Assumptions

A solid sphere of radius $R$ carries a uniform volume charge density $\sigma$ with units $\mathrm{C/m^3}$. A spherical cavity of radius $r$ is removed from this sphere. The center of the cavity is displaced from the center of the original sphere by a vector $\mathbf{a}$ with magnitude $a$ and units $\mathrm{m}$. The permittivity of free space is $\varepsilon_0$ with units $\mathrm{F/m}$.

The task is to determine the electric field $\mathbf{E}$ at arbitrary points inside the cavity. The system is electrostatic, charges are fixed in space, and edge effects at the boundary of the cavity are included through exact field superposition.

The physical model treats the cavity as a region where the charge density is reduced from $\sigma$ to zero, implemented by superposing a uniformly charged sphere of density $-\sigma$ occupying the cavity volume.

Physical Principles

The electric field of a continuous charge distribution is governed by the principle of superposition,

$\mathbf{E} = \mathbf{E}_1 + \mathbf{E}_2,$

where each contribution is computed independently.

For a uniformly charged solid sphere of constant volume charge density $\sigma$, Gauss’s law in integral form,

$\oint \mathbf{E}\cdot d\mathbf{S} = \frac{Q_{\text{enc}}}{\varepsilon_0},$

implies that the electric field at a point with position vector $\mathbf{r}$ measured from the center of the sphere is

$\mathbf{E}(\mathbf{r}) = \frac{\sigma}{3\varepsilon_0}\mathbf{r},$

valid for points inside the sphere.

A cavity is represented by superposition of a sphere of charge density $-\sigma$ centered at the cavity center.

Derivation

Let the origin be at the center of the full sphere, and let the position vector of a field point inside the cavity be $\mathbf{r}$. Let the center of the cavity be at position vector $\mathbf{a}$.

The full uniformly charged sphere produces inside-field

$\mathbf{E}_1(\mathbf{r}) = \frac{\sigma}{3\varepsilon_0}\mathbf{r}.$

The cavity is modeled as a second sphere of charge density $-\sigma$ centered at $\mathbf{a}$. For points inside this cavity sphere, the field due to this fictitious sphere is

$\mathbf{E}_2(\mathbf{r}) = \frac{-\sigma}{3\varepsilon_0}(\mathbf{r} - \mathbf{a}).$

The total field inside the cavity is the sum

$\mathbf{E}(\mathbf{r}) = \mathbf{E}_1(\mathbf{r}) + \mathbf{E}_2(\mathbf{r}).$

Substituting the expressions gives

$\mathbf{E}(\mathbf{r}) = \frac{\sigma}{3\varepsilon_0}\mathbf{r} - \frac{\sigma}{3\varepsilon_0}(\mathbf{r} - \mathbf{a}).$

Expanding the second term yields

$\mathbf{E}(\mathbf{r}) = \frac{\sigma}{3\varepsilon_0}\mathbf{r} - \frac{\sigma}{3\varepsilon_0}\mathbf{r} + \frac{\sigma}{3\varepsilon_0}\mathbf{a}.$

The dependence on $\mathbf{r}$ cancels exactly, leaving

$\mathbf{E}(\mathbf{r}) = \frac{\sigma}{3\varepsilon_0}\mathbf{a}.$

Result

The electric field at every point inside the cavity is uniform and given by

$\boxed{\mathbf{E} = \frac{\sigma}{3\varepsilon_0}\mathbf{a}}.$

No numerical substitution is required because the result is expressed entirely in terms of the given parameters $\sigma$, $\varepsilon_0$, and $\mathbf{a}$. The magnitude of the field is

$E = \frac{\sigma a}{3\varepsilon_0},$

with direction from the center of the original sphere toward the center of the cavity.

Sanity Checks

Dimensional consistency follows from $\sigma/\varepsilon_0$ having units $\mathrm{(C/m^3)/(C/(V\cdot m)) = V/m^2}$ multiplied by $a$ with units $\mathrm{m}$, producing $\mathrm{V/m}$ for the electric field.

In the limit $a \to 0$, the cavity becomes concentric with the charged sphere, and the field vanishes everywhere inside, consistent with symmetry.

In the limit of small charge density $\sigma \to 0$, the field disappears linearly with the source strength, matching electrostatic linearity.

A sign error most easily arises in the superposition step for the cavity contribution, where forgetting the minus sign for the removed charge would incorrectly produce a position-dependent field rather than a uniform one.