A point charge \(Q\) is a distance \(d\) from the centre \(O\) of an earthed conducting sphere of radius \(a\). Show that a charge \(-Q a / d\) at a distance \(a^{2} / d\) from \(\mathrm{O}\) along \(\mathrm{OQ}\) is a satisfactory image charge. Find the ratio of the maximum to the minimum surface density of charge induced on the sphere.

Diving into the universe of electrostatics, we encounter a fundamental concept: electric potential, a measure of the potential energy per unit charge at a specific point in an electric field. Imagine you're toying with a tiny charged ball in an electric playground, and you want to move it around. The electric potential tells you how much work you'd need to push this ball against the electrostatic forces at play.

Explaining this in a practical scenario, a point charge creates an electrostatic playground extending through space. If you bring another charge into this playground, it feels a tug, pulling or pushing it due to the existing electric field. The electric potential at any point is then calculated using the formula \(V = \frac{kQ}{r}\), where \(k\) is Coulomb's constant, \(Q\) represents the charge creating the field, and \(r\) is the distance from \(Q\) to the point of interest.

We see electric potential at play in the solution of the textbook problem. An image charge provides a way to simplify a complex electrostatic scenario by using a 'mirror charge', here placed at a specified distance and ensuring the conducting sphere has zero potential, as dictated by it being earthed. Thus, the sphere equilibrates to the same electric potential as the ground, and this potential is conceptually the electric 'altitude' at which it is comfortably sitting—zero in this case.

Surface charge density is akin to a crowd's density at a concert—how closely packed is the electric 'audience' on the surface of a material? Rigorously speaking, it's the amount of electric charge per unit area on the surface of a conductor. In our problem, it's crucial as it reflects how the sphere's charge is reshuffling itself in response to the nearby point charge.

To calculate it, imagine taking a tiny, imperceptibly small magnifying glass and looking so closely at the surface that you can count individual charges in a teeny-weeny area. The ratio of the total charge to the area you're observing gives the surface charge density, \(\sigma \). Mathematically, if \( Q_{surface} \) is the charge over a small patch of area \( A \), then \( \sigma = \frac{Q_{surface}}{A} \).

In our scenario, the surface charge density changes from point to point on the sphere due to the presence of the point charge and its image charge. The region closest to the point charge would have a higher concentration of charge, akin to fans pushing towards the stage at a concert, while the opposite side would be less densely packed.

Anchoring many problems in electrostatics, Gauss' law is a powerful tool encapsulating the relationship between electric charges and the resulting electric field. It is the electrical equivalent of placing a net around a school of fish to count them, linking the net's mesh size (electric field) to the quantity of fish (charge) inside.

Gauss' law formally states that the electric flux out of any closed surface is equal to the net charge enclosed divided by the vacuum permittivity, with the equation: \(\Phi = \frac{Q_{enc}}{\epsilon_0}\). Consider \(\epsilon_0\) the 'net tightness' (a constant value determining how the electric field relates to the enclosed charge) and \(\Phi\) as the total fish count (electric flux).

Applying Gauss' law can be thought of as creating an imaginary bubble around our conductor—the 'Gaussian surface'. Any imbalance in charge within this bubble affects the electric field outside it. In our textbook example, we use this principle to deduce the surface charge density at different points on the sphere due to the image charge. By considering a small patch on the sphere's surface, the local electric field dictates the crowd density (surface charge density) through Gauss' law.