Kvant Physics Problem 6

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Problem

Find the charge of a grounded conducting sphere of radius $r$ if a point charge $q$ is located at a distance $R$ from its center.

2nd All-Union Physics Olympiad

Setup and Assumptions

A conducting sphere of radius $r$ is connected to ground. A point charge $q$ is placed outside the sphere at a distance $R$ from the center, with $R>r$.

The unknown quantity is the total charge $Q$ induced on the grounded sphere.

The conductor is assumed to be ideal, electrostatic equilibrium is established, the surrounding medium is vacuum with permittivity $\varepsilon_0$, and gravitational effects are neglected. The grounding wire is assumed ideal, so the electric potential of the sphere is maintained at zero. The point charge is treated as a mathematical point charge.

Physical Principles

The solution is based on the method of images.

For a grounded conducting surface, the electrostatic potential on the surface must satisfy

$$\varphi=0.$$

The electric field outside the conductor can be reproduced by replacing the conductor with suitable image charges located inside the sphere, provided that the resulting potential satisfies the boundary condition on the spherical surface.

For a point charge $q$ located at distance $R$ from the center of a grounded conducting sphere of radius $r$, the image solution consists of a single image charge

$$q'=-q\frac{r}{R},$$

placed on the line joining the center to the real charge at distance

$$a=\frac{r^2}{R}$$

from the center.

The uniqueness theorem guarantees that this image configuration reproduces the actual electric field outside the sphere.

Derivation

Choose the center of the sphere as the origin and place the real charge on the positive $z$ axis. The image charge is then located on the same axis inside the sphere.

The potential at an exterior point is

$$\varphi(\mathbf r) = \frac{1}{4\pi\varepsilon_0} \left( \frac{q}{|\mathbf r-\mathbf R|} + \frac{q'}{|\mathbf r-\mathbf a|} \right).$$

The values

$$q'=-q\frac{r}{R}, \qquad a=\frac{r^2}{R},$$

are chosen so that $\varphi=0$ for every point on the spherical surface $|\mathbf r|=r$.

The actual charge distribution induced on the conductor produces exactly the same external field as the image charge. To find the total induced charge, consider a Gaussian surface enclosing the sphere and lying entirely inside the region $r<\rho<R$, so that it surrounds the conductor but not the real charge $q$.

In the image problem, the electric field in this region is produced by the image charge alone. By Gauss's law, the total charge enclosed by such a surface equals the image charge:

$$Q=q'.$$

Hence the total charge induced on the grounded sphere is

$$Q=-q\frac{r}{R}.$$

This is the required charge of the sphere.

Result

The charge acquired by the grounded conducting sphere is

$$Q=-q\frac{r}{R}.$$

Since the problem gives only symbolic parameters, no numerical substitution can be performed.

The final answer is

$$\boxed{Q=-q\frac{r}{R}}.$$

Its unit is the same as the unit of $q$, namely coulombs.

Sanity Checks

The ratio $r/R$ is dimensionless, so the expression has the same units as $q$. Dimensional consistency is satisfied.

If the external charge is moved very far away, $R\to\infty$, then

$$Q=-q\frac{r}{R}\to 0.$$

A distant charge produces only a weak disturbance, so the sphere acquires a vanishingly small net charge.

If the charge approaches the sphere, $R\to r^+$, then

$$Q\to -q.$$

The induced charge becomes nearly equal in magnitude and opposite in sign to the external charge, which is physically reasonable.

The sign is negative for a positive external charge. Electrons flow from the ground onto the sphere until the sphere remains at zero potential, so the sphere acquires negative charge.

The step most susceptible to error is the relation

$$q'=-q\frac{r}{R}.$$

A missed minus sign would predict a positive induced charge for a positive external charge, contradicting the physical behavior of a grounded conductor.