Kvant Physics Problem 180

Verified: yes
Verdicts: PASS + PASS
Solve time: 5m14s
Source on kvant.digital

Problem

The problem was mistakenly published under number 182 in the journal.

In a uniform electric field of strength $E$, a metal sphere is placed. It is known that the surface charge density at the “pole” of the sphere at point $A$ (Fig. 3) is equal to $\sigma_0$. Determine the surface charge density at the point $B$, whose direction from the center of the sphere makes an angle $\alpha$ with the direction of the external electric field.

Figure 3

Yu. A. Dreizin

Setup and Assumptions

A perfectly conducting isolated metal sphere of radius $R$ is placed in a uniform external electric field of magnitude $E$ directed along a fixed axis. The conductor reaches electrostatic equilibrium, so the electric field inside the metal is zero and all excess charge resides on the surface. The induced surface charge density depends only on the polar angle $\theta$ measured from the direction of the external field.

Point $A$ is located at the “pole” of the sphere where $\theta = 0$, and the surface charge density there is given as $\sigma_0$ with units $\mathrm{C/m^2}$. Point $B$ lies on the surface at polar angle $\theta = \alpha$, and the quantity to determine is the surface charge density $\sigma(B)$ in $\mathrm{C/m^2}$.

Edge effects beyond the sphere are neglected, the surrounding medium is vacuum with permittivity $\varepsilon_0$, and the system is treated within classical electrostatics.

Physical Principles

In electrostatic equilibrium, the electric field just outside a conductor is related to the surface charge density by the boundary condition

$$E_n = \frac{\sigma}{\varepsilon_0},$$

where $E_n$ is the normal component of the electric field immediately outside the surface.

The electrostatic potential $V$ in the region outside the conductor satisfies Laplace’s equation

$$\nabla^2 V = 0.$$

For a conducting sphere in a uniform external field, the unique solution consistent with axial symmetry and the boundary condition of constant potential on the conductor surface leads to a surface charge distribution proportional to $\cos\theta$.

The standard electrostatic solution yields the normal field at the surface of a conducting sphere in a uniform field $E$ as

$$E_n(\theta) = 3E\cos\theta.$$

Combining this with the boundary condition between field and surface charge density determines $\sigma(\theta)$.

Derivation

At the surface of a conductor in electrostatic equilibrium, the normal electric field just outside the surface is related to the surface charge density by

$$\sigma(\theta) = \varepsilon_0 E_n(\theta).$$

For a conducting sphere placed in a uniform external field of magnitude $E$, the solution of Laplace’s equation with axial symmetry gives the normal component of the electric field on the surface as

$$E_n(\theta) = 3E\cos\theta.$$

Substituting this into the boundary relation yields

$$\sigma(\theta) = \varepsilon_0 \cdot 3E\cos\theta,$$

so

$$\sigma(\theta) = 3\varepsilon_0 E \cos\theta.$$

At the pole $A$, where $\theta = 0$, the surface charge density is

$$\sigma_0 = 3\varepsilon_0 E \cos 0 = 3\varepsilon_0 E.$$

Eliminating $3\varepsilon_0 E$ in favor of $\sigma_0$ gives

$$\sigma(\theta) = \sigma_0 \cos\theta.$$

At point $B$, where $\theta = \alpha$, the surface charge density becomes

$$\sigma(B) = \sigma_0 \cos\alpha.$$

Result

The surface charge density at point $B$ is

$$\sigma(B) = \sigma_0 \cos\alpha.$$

Substituting units, $\sigma_0$ is in $\mathrm{C/m^2}$ and $\cos\alpha$ is dimensionless, so

$$\sigma(B) = \sigma_0 \cos\alpha \quad \mathrm{C/m^2}.$$

Sanity Checks

At $\alpha = 0$, the expression gives $\sigma(B) = \sigma_0$, matching the definition of the pole value. At $\alpha = \frac{\pi}{2}$, the result becomes $\sigma(B) = 0$, consistent with the symmetry point where the normal component of the external field vanishes. At $\alpha = \pi$, the charge density becomes $\sigma(B) = -\sigma_0$, reflecting charge accumulation of opposite sign on the far pole.

Dimensional consistency holds because $\cos\alpha$ is dimensionless and the result retains units of surface charge density. The linear dependence on $E$ hidden in $\sigma_0 = 3\varepsilon_0 E$ ensures correct proportionality to the applied field strength, and any sign error would most easily arise from reversing the direction chosen for $\theta = 0$ relative to the external field direction.