A solid sphere \(25 \mathrm{cm}\) in radius carries \(14 \mu \mathrm{C},\) distributed uniformly throughout its volume. Find the electric field strength (a) \(15 \mathrm{cm},\) (b) \(25 \mathrm{cm},\) and (c) \(50 \mathrm{cm}\) from its center.

Understanding how electric charge is distributed is crucial for comprehending many principles of electrostatics. Regarding a solid sphere with a uniform electric charge distribution, like in our exercise, this means the charge density is the same throughout the volume of the sphere.

When a charge is uniformly distributed, it simplifies the calculations for finding electrical properties. For instance, to find the charge within any region inside the sphere, we think of slicing the sphere into concentric shells. The calculation of the charge within a certain radius is proportional to the volume of that sphere segment. This is because with uniform distribution, a larger volume contains proportionally more charge. Considering our example, the solid sphere of radius 25 cm carries a charge of 14 μC uniformly; we can calculate the charge inside a smaller radius using the volumes ratio.

Electrostatics is the branch of physics that deals with the study of stationary electric charges or charges at rest. A key aspect of electrostatics is understanding the effects that charges have on each other. For instance, charges produce an electric field, which can exert a force on other charges within the field.

In the context of our problem, we are analyzing the electrostatic effect of a uniformly charged solid sphere. For a solid sphere with a uniform charge distribution, the electric field varies depending on where you measure it — whether it be inside, at the surface, or outside of the sphere. Such concepts are pivotal in grasping how electric fields behave in different scenarios.

Gauss's law for electricity is a powerful tool derived from Coulomb's law that provides a method to calculate the electric field generated by various charge distributions. It is expressed as \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \] where \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is the differential area vector, \( Q_{\text{enc}} \) is the total charge enclosed within the surface, and \( \varepsilon_0 \) is the permittivity of free space.

Gauss's law is particularly useful for calculating the electric field of symmetric charge distributions, such as our solid sphere. It tells us that for a point inside the sphere, only the charge enclosed by a Gaussian surface matters. Thus, a spherical Gaussian surface concentric with our charged sphere can simplify our problem significantly.

Calculating the electric field involves determining the force a charge would experience per unit charge at a given point in space. Using Coulomb's law and the principles of superposition, we can calculate the electric fields for complex charge distributions. In the context of our solid sphere, we apply different methods for points inside, on the surface of, or outside the sphere.

For instance, at a point inside the sphere, the electric field strength can be found using the formula \[ E=k \frac{q_{\text{inside}}}{r^2} \], with \( q_{\text{inside}} \) being the charge within the Gaussian surface at that radius and \( r \) being the distance from the center of the sphere to the point. On its surface or outside, the entire charge can be treated as if it were concentrated at the center of the sphere. Through these calculations, as shown in the exercise, students gain a deeper understanding of electric fields and electrostatic principles.