A uniformly charged thin spherical shell of radius \(R\) carries uniform surface charge density of \(\sigma\) per unit area. It is made of two hemispherical shells, held together by pressing them with force \(F\) (see figure). \(F\) is proportional toA) \(\frac{1}{\varepsilon_{0}} \sigma^{2} \mathrm{R}^{2}\) B) \(\frac{1}{\varepsilon_{0}} \sigma^{2} \mathrm{R}\) C) \(\frac{1}{\varepsilon_{0}} \frac{\sigma^{2}}{R}\) D) \(\frac{1}{\varepsilon_{0}} \frac{\sigma^{2}}{R^{2}}\)

Understanding the concept of surface charge density is fundamental when studying electric fields and forces on charged surfaces. Surface charge density, represented by the symbol \( \sigma \), is defined as the amount of electric charge per unit area on a surface. In the case of our exercise, we imagine a uniformly charged thin spherical shell. This uniformity means that every tiny patch of the shell's surface carries an identical amount of charge.

For a given charge \( Q \) distributed over a surface area \( A \), the surface charge density \( \sigma \) is calculated using the formula: \[ \sigma = \frac{Q}{A} \]. Now, when we think about the pressure that a charged object experiences in space, this density helps us understand the distribution of forces across the object's surface. The idea is simple—the greater the surface charge density, the stronger the repelling forces between similar charges in close proximity.

The electric field is a fundamental concept in electromagnetism, describing the force that a charged particle would experience at any point in space. When dealing with a charged shell, the electric field plays a crucial role in determining the forces at play.

In our scenario, the electric field just outside the surface of the charged shell with charge density \( \sigma \) is given by \( E = \frac{\sigma}{2\varepsilon_0} \), where \( \varepsilon_0 \) is the vacuum permittivity. It's crucial to note that the shell's interior bears no electric field—this is a result of the symmetry in a spherical charge distribution, as per Gauss's law. The field right outside the shell, however, is comparable to that created by an equivalent point charge at a distance larger than the shell's radius.

Understanding the distribution of the electric field around the charged shell is essential because it directly affects the electric pressure exerted on the shell’s surface. The electric field essentially dictates how charges interact with one another, influencing the force and pressure experienced by the shell.

Electric pressure is a less commonly discussed concept, but in the context of charged surfaces, it's incredibly important. It's the pressure exerted by the electric field on a charged surface, and it can be thought of as the force per unit area. Electric pressure is caused by the repulsion or attraction between charged particles.

For our uniformly charged shell, the electric pressure is derived from the electric field, using the formula \( P = \frac{E^2}{2\varepsilon_0} \). By substituting the earlier found electric field value, we calculate the pressure to be \( P = \frac{\sigma^2}{4\varepsilon_0^2} \). This pressure causes a force that acts to push the two hemispherical shells apart.

To compute the total force \( F \) due to this pressure on one of the hemispheres, we multiply the electric pressure by the area over which it acts, \( A = 2\pi R^2 \). This results in the force formula: \[ F = P \cdot A = \frac{\sigma^2}{4\varepsilon_0^2} \cdot 2\pi R^2 \]. Simplifying this expression, we link force to the original parameters of charge density and the radius of the shell.

Remember, this exercise isn't just about calculations—it teaches us about the fascinating ways in which charged particles interact and the forces they exert on one another, all based on fundamental physical laws.