Kvant Physics Problem 180

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Problem

It was mistakenly published in the journal under number 182.

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

Figure 3

Yu. A. Dreizin

Setup and Assumptions

A perfectly conducting sphere of radius $R$ is placed in an initially uniform electrostatic field of magnitude $E$, directed along a fixed axis. After electrostatic equilibrium is reached, the electric field inside the conductor vanishes and surface charges redistribute on the sphere. The surface charge density on the sphere is denoted by $\sigma(\theta)$, where $\theta$ is the angle between the radius vector to a surface point and the direction of the external field.

Point $A$ corresponds to the “pole” aligned with the external field direction, so $\theta=0$ there and $\sigma(0)=\sigma_0$. Point $B$ lies on the sphere in the direction making an angle $\alpha$ with the field, so its surface charge density is $\sigma(\alpha)$, which is required.

The surrounding medium is vacuum with permittivity $\varepsilon_0$. Edge effects, polarization of the medium, and any non-electrostatic forces are neglected.

Physical Principles

In electrostatic equilibrium inside a conductor, the electric field satisfies $\mathbf{E}=0$, and the conductor surface is an equipotential.

The discontinuity of the normal component of the electric field at a conductor surface gives the surface charge density through

$$\sigma = \varepsilon_0 E_n,$$

where $E_n$ is the outward normal component of the electric field just outside the conductor.

The electrostatic potential in the exterior region satisfies Laplace’s equation. For a conducting sphere in a uniform external field aligned with the polar axis, the axially symmetric solution takes the form

$$\Phi(r,\theta) = -E r \cos\theta + \frac{A \cos\theta}{r^2},$$

where the constant $A$ is determined from the boundary condition that the sphere surface is an equipotential.

The electric field is obtained from the potential as

$$\mathbf{E} = -\nabla \Phi.$$

Derivation

The potential outside the sphere is written as

$$\Phi(r,\theta) = -E r \cos\theta + \frac{A \cos\theta}{r^2}.$$

The conductor surface at $r=R$ must have constant potential, independent of $\theta$. Substituting $r=R$ gives

$$\Phi(R,\theta) = -E R \cos\theta + \frac{A \cos\theta}{R^2}.$$

For this expression to be independent of $\theta$, the coefficient of $\cos\theta$ must vanish, which yields

$$-E R + \frac{A}{R^2} = 0,$$

so

$$A = E R^3.$$

The potential becomes

$$\Phi(r,\theta) = -E r \cos\theta + E R^3 \frac{\cos\theta}{r^2}.$$

The radial component of the electric field is

$$E_r = -\frac{\partial \Phi}{\partial r}.$$

Differentiating term by term gives

$$\frac{\partial \Phi}{\partial r} = -E \cos\theta - 2E R^3 \frac{\cos\theta}{r^3}.$$

Thus

$$E_r = E \cos\theta + 2E R^3 \frac{\cos\theta}{r^3}.$$

Evaluating at the surface $r=R$ produces

$$E_r(R,\theta) = E \cos\theta + 2E \cos\theta = 3E \cos\theta.$$

The surface charge density follows from the boundary condition

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

At the pole $A$ where $\theta=0$,

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

Solving for the field strength gives

$$E = \frac{\sigma_0}{3\varepsilon_0}.$$

Substituting into the general expression yields

$$\sigma(\theta) = 3\varepsilon_0 \cdot \frac{\sigma_0}{3\varepsilon_0} \cos\theta = \sigma_0 \cos\theta.$$

For point $B$ at angle $\alpha$, this becomes

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

Result

The surface charge density at point $B$ is

$$\sigma_B = \sigma_0 \cos\alpha.$$

No numerical substitution is required beyond the given symbolic parameters, so the result remains

$$\boxed{\sigma_B = \sigma_0 \cos\alpha}.$$

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 yields $\sigma_B=0$, consistent with the vanishing normal field at the equator of a conducting sphere in a uniform external field.

At $\alpha=\pi$, the expression gives $\sigma_B=-\sigma_0$, corresponding to charge accumulation of opposite sign on the opposite pole.

Dimensional consistency follows since both $\sigma_B$ and $\sigma_0$ carry units of $\mathrm{C,m^{-2}}$, while $\cos\alpha$ is dimensionless. The only sensitive step where an error would propagate is the evaluation of the radial derivative of $\Phi$, since a missing factor of $2$ in the derivative of $r^{-2}$ would destroy the factor $3$ in the surface field and alter the proportionality between $\sigma_0$ and $E$.