Kvant Physics Problem 173

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Problem

In the journal it was mistakenly published under number 163.

In a uniformly charged sphere of radius $R$ there is a spherical cavity of radius $r$, whose center is located 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 uniformly charged solid sphere of radius $R$ carries a constant volume charge density $\sigma$ with SI units $\mathrm{C,m^{-3}}$. Inside this sphere a spherical cavity of radius $r$ is removed. The cavity center is displaced from the sphere center by a fixed vector $\mathbf{a}$ with magnitude $a$ and units $\mathrm{m}$.

The task is to determine the electric field $\mathbf{E}$ at arbitrary points inside the cavity.

The system is treated within classical electrostatics. The charge distribution is static. The medium is vacuum with permittivity $\varepsilon_0$ in units $\mathrm{F,m^{-1}}$. Edge effects at the boundary of the removed material are included exactly through superposition.

Physical Principles

The electric field obeys Gauss’s law in differential form

$\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0},$

where $\rho$ is the volume charge density.

For a uniformly charged full sphere of density $\sigma$, the field at a point with position vector $\mathbf{r}$ measured from its center, provided $\lvert \mathbf{r} \rvert < R$, is

$\mathbf{E}_{\text{sphere}} = \frac{\sigma}{3\varepsilon_0}\mathbf{r}.$

The principle of superposition applies, so the field of a composite charge distribution equals the vector sum of the fields of its parts.

Derivation

Let the origin be at the center of the full sphere. Let $\mathbf{r}$ denote the position vector of an arbitrary point inside the cavity.

The cavity is represented as a superposition of a full sphere of charge density $\sigma$ centered at $\mathbf{a}$ and radius $r$, added with negative sign to the original full sphere.

The electric field of the original uniformly charged sphere at point $\mathbf{r}$ is

$\mathbf{E}_1 = \frac{\sigma}{3\varepsilon_0}\mathbf{r}, \qquad \lvert \mathbf{r} \rvert < R.$

The field of the removed spherical region, treated as a sphere of density $\sigma$ centered at $\mathbf{a}$, evaluated at the same point $\mathbf{r}$, is

$\mathbf{E}_2 = \frac{\sigma}{3\varepsilon_0}(\mathbf{r} - \mathbf{a}), \qquad \lvert \mathbf{r} - \mathbf{a} \rvert < r.$

Since this charge is removed, its contribution enters with a negative sign. The total field inside the cavity is therefore

$\mathbf{E} = \mathbf{E}_1 - \mathbf{E}_2.$

Substituting the expressions gives

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

Expanding the parentheses yields

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

Cancellation of the position-dependent terms leads to

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

The resulting field is independent of $\mathbf{r}$, which means it is uniform throughout the cavity.

Result

The electric field at any point inside the cavity is

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

Its magnitude is

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

In SI units, $\sigma$ is measured in $\mathrm{C,m^{-3}}$, $a$ in $\mathrm{m}$, and $\varepsilon_0$ in $\mathrm{F,m^{-1}}$, giving $\mathbf{E}$ in $\mathrm{N,C^{-1}}$ or equivalently $\mathrm{V,m^{-1}}$.

Sanity Checks

The expression depends linearly on the displacement vector $\mathbf{a}$, so when the cavity center coincides with the sphere center, $a=0$, the field vanishes everywhere inside the cavity, matching the symmetry of a full uniformly charged sphere.

Dimensional analysis gives $\sigma a / \varepsilon_0$, which reduces to $\mathrm{C,m^{-3}}\cdot \mathrm{m} / \mathrm{F,m^{-1}} = \mathrm{C,m^{-2}} / \mathrm{F,m^{-1}} = \mathrm{N,C^{-1}}$, consistent with electric field units.

The final expression contains no dependence on $r$, the cavity radius, which is consistent with the fact that only the removed sphere’s center contributes to the uniform field inside the cavity. A sign error would most likely occur when subtracting the displaced sphere contribution, which would incorrectly reverse the direction of $\mathbf{E}$ relative to $\mathbf{a}$.