A spherical shell \(30 \mathrm{cm}\) in diameter carries \(85 \mu \mathrm{C}\) distributed uniformly over its surface. A \(1.0-\mu \mathrm{C}\) point charge is located at the shell's center. Find the electric field strength (a) \(5.0 \mathrm{cm}\) from the center and (b) \(45 \mathrm{cm}\) from the center. (c) How would your answers change if the charge on the shell were doubled?

Understanding Gauss' Law is essential to grasp the behavior of electric fields in different scenarios. It states that the electric flux through a closed surface, known as a Gaussian surface, is directly proportional to the enclosed electric charge. In mathematical terms, it is expressed as \[ \Phi_E = \frac{Q_{enc}}{\epsilon_0} \] where \( \Phi_E \) is the electric flux, \( Q_{enc} \) is the enclosed charge, and \( \epsilon_0 \) is the vacuum permittivity, a fundamental constant.

In practical terms, for a spherical shell like the one described in the exercise, Gauss' Law implies that the electric field inside the shell is only affected by charges located within the Gaussian surface, not by the charges on or outside it. That's why the electric field inside the sphere due to the shell's charge is zero - it's completely external to any Gaussian surface we might draw around a point inside the shell. However, the point charge at the center does contribute to the electric field inside the shell, as it is enclosed by the surface.

To calculate the electric field, one would typically choose a spherical Gaussian surface concentric with the shell and passing through the point of interest. The symmetry of the problem greatly simplifies the calculations, as it allows us to state that the electric field has the same magnitude at every point on the Gaussian surface and is directed radially.

Coulomb's constant, denoted by \(k\), plays a pivotal role in the equation that describes the force between two point charges. It is a proportionality constant in Coulomb's law and has a value of approximately \(8.99 \times 10^9 \mathrm{N {m}^2 {C}^{-2}}\). This constant helps us quantify the strength of the electric field and force between point charges.

The formula to calculate the electric field \(E\) due to a single point charge is given by \[ E = \frac{kQ}{r^2} \] where \(Q\) is the charge and \(r\) is the distance from the charge to the point at which you're calculating the field. This equation is the direct application of Coulomb's law to electric field strength and is crucial for solving many electrostatic problems, including the exercise we are reviewing.

When dealing with problems involving point charges, always remember that the electric field is directly proportional to the charge and inversely proportional to the square of the distance from the charge. This relationship is why the electric field inside the sphere solely depends on the point charge at the center and not on the charge distributed over the shell's surface.

The concept of an electric field due to a point charge is foundational in understanding electrostatics. As mentioned earlier, the electric field \(E\) at a distance \(r\) from a point charge \(Q\) is calculated using the formula \[ E = \frac{kQ}{r^2} \]. This equation implies that the electric field decreases with the square of the distance from the charge, indicating that fields diminish rapidly as one moves away from the source charge.

For the 1.0-\(\mu\mathrm{C}\) point charge at the center of the shell in our exercise, we directly apply this formula to find the electric field strength 5.0 cm away. Despite other charges present, such as on the spherical shell, this central charge is solely responsible for the electric field at that particular point inside the shell. Outside the shell, the total charge creates the electric field, treating the shell and its enclosed charge as a single point charge at the center.

It is this concept that allows us to compute the electric field at various points and understand how it is influenced by the geometry and distribution of charges. Always bear in mind that the electric field vector points away from positive charges and towards negative ones, a fact that helps in visualizing electric field lines and understanding the forces that a charge will experience in the presence of others.