If some apparatus is to be operated at \(90 \mathrm{kV}\), what is the smallest external radius of curvature which should be permitted? (Dielectric strength of air \(=30 \mathrm{kV} \mathrm{cm}^{-1} .\) ) If smaller radii are unavoidable, what is a possible solution to the difficulty?

The concept of dielectric strength is critical when designing electrical apparatus. It is defined as the maximum electric field that a material can withstand without undergoing electrical breakdown. In simpler terms, it's the electrical threshold at which an insulating material starts to conduct electricity, leading to a potential failure known as an electrical breakdown. No material is a perfect insulator, and when the electric field within a material exceeds its dielectric strength, the electrons in the material become sufficiently energized to free themselves from their atomic bonds and start to flow, which can cause damage or even destruction to the material or associated electrical components. In the textbook exercise, it's stated that the dielectric strength of air is 30 kV/cm, meaning if an electric field exceeds this limit, the air could begin conducting electricity, causing a hazardous arc or spark.

For an apparatus operating at 90 kV, it's crucial to ensure the external radius of curvature on any part of the device is large enough to prevent the electric field from exceeding this dielectric strength of air. Hence, when regions of smaller radius of curvature appear unavoidable, replacing air with another material that has higher dielectric strength is imperative. Examples of such materials include insulating oils, ceramics, and various gases like sulfur hexafluoride (SF6) under pressure.

Coulomb's law plays a fundamental role in understanding electric fields around charged objects. It describes the force between two point charges as directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The mathematical expression is given as:
\[ F = k \frac{{|q_1 q_2|}}{{r^2}} \
where \( F \) is the electrostatic force, \( k \) is Coulomb's constant (approx. \( 8.99 \times 10^9 N m^2/C^2 \)), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges. In the solved example, Coulomb's law is used to determine the electric field at a distance R from the center of a spherical conductor, which could represent a scaled-down, simplified model for the apparatus in question. According to this law, the electric field decreases with the square of the distance from the charge, emphasizing why increasing the radius of the apparatus can significantly lower the electric field intensity on its surface to avoid surpassing the dielectric strength of air.

Electric breakdown occurs when an insulating material becomes electrically conductive due to the presence of an extremely high electric field, resulting in an uncontrolled flow of current. This phenomenon essentially transforms a non-conductive material into a conductive one, at least temporarily. When the electric field within a material (like the air in our example) surpasses its dielectric strength, electrons are pulled away from their atoms, starting a cascade of collisions and an avalanche of charge carriers which leads to a conductive pathway through the material, often accompanied by visible sparks or arcs.

Electric breakdown is not just limited to air; it can occur in solids and liquids as well and is a prime concern in high-voltage applications. Preventing electric breakdown in equipment operating at high voltages, like the 90 kV apparatus in our example, is crucial as it could cause severe damage, fires, short circuits, or system failures. Using materials with higher dielectric strength around components where high electric fields are expected helps mitigate this risk profoundly.

The charge-voltage relationship for a spherical conductor is a direct outcome of Coulomb's law and explains how the potential (voltage) on a charged spherical conductor’s surface is related to its stored electric charge. The relationship is defined by the equation:
\[ V = \frac{kQ}{R} \
Here, \( V \) is the electric potential or voltage, \( k \) is Coulomb's constant, \( Q \) is the charge on the conductor, and \( R \) is the radius of the conductor. When this equation is applied, it directly links the voltage across a sphere to its radius, given a certain amount of charge. Understanding this relationship allows us to manipulate the radius and charge to control the voltage and vice versa. In our solved exercise, the electrical potential (90 kV) is linked with the charge on the conductor to determine the minimum required radius to keep the electric field below the dielectric strength of air, thereby preventing electric breakdown. If the operating voltage is fixed, as is common in many practical situations, ensuring proportional adjustments in the sphere’s radius or the charge becomes necessary to maintain the electric field within safe limits.