Question

A uniform sphere has a radius \(R\) and a total charge \(+Q\), uniformly distributed throughout its volume. It is surrounded by a thick spherical shell carrying a total charge \(-Q,\) also uniformly distributed, and having an outer radius of \(2 R\). What is the electric field as a function of \(R ?\)

Short answer

Based on the given solution, describe the electric field inside the sphere, between the sphere and the shell, and outside the shell. Inside the sphere (r < R), the electric field is proportional to the radial distance from the center of the sphere and is given by: $$ E(r) = \frac{1}{\varepsilon_0} \cdot \frac{Q}{4\pi r^2} \cdot \frac{r^3}{R^3}. $$ Between the sphere and the shell (R < r < 2R), the electric field is calculated by considering the charges of both the sphere and part of the shell. It is given by: $$ E(r) = \frac{1}{\varepsilon_0} \cdot \frac{Q-Q(r)}{4\pi r^2}. $$ Outside the shell (r > 2R), the electric field is zero since the total enclosed charge for the sphere with positive charge (+Q) and the shell with negative charge (-Q) is zero: $$ E(r) = 0. $$

Step-by-step solution

Case 1: \(r

For points inside the sphere, we consider a Gaussian surface in the form of a sphere with radius \(r<R\). The electric field at a distance \(r\) from the center of the sphere is radial and has the same magnitude at every point on the Gaussian surface. By applying Gauss's Law, we can find the electric field: $$ E(r) \cdot 4\pi r^2 = \frac{1}{\varepsilon_0} \cdot \frac{Q(r)}{V} \cdot \frac{4}{3}\pi r^3, $$ where \(Q(r)\) is the total charge inside the Gaussian sphere of radius \(r\) and \(V = \frac{4}{3}\pi R^3\) is the volume of the uniformly charged sphere. Solving for the electric field: $$ E(r) = \frac{1}{\varepsilon_0} \cdot \frac{Q(r)}{4\pi r^2}. $$ Since the charge \(Q\) is uniformly distributed within the sphere, the charge enclosed within the Gaussian sphere can be found using: $$ Q(r) = Q \cdot \frac{r^3}{R^3}, $$ Substituting this into the electric field equation, we get: $$ E(r) = \frac{1}{\varepsilon_0} \cdot \frac{Q}{4\pi r^2} \cdot \frac{r^3}{R^3}. $$

Case 2: \(R

For points located between the sphere and the outer radius of the shell, we again consider a Gaussian surface in the form of a sphere with radius \(r\). In this case, the Gaussian surface encloses the entire charged sphere and only a part of the shell. Applying Gauss's Law, we find the electric field in the same way as in case 1: $$ E(r) \cdot 4\pi r^2 = \frac{1}{\varepsilon_0} \cdot \frac{Q-Q(r)}{4\pi r^2}, $$ Solving for the electric field: $$ E(r) = \frac{1}{\varepsilon_0} \cdot \frac{Q-Q(r)}{4\pi r^2}. $$ Here, \(Q(r)\) is the amount of charge enclosed in the Gaussian sphere and can be written as: $$ Q(r) = -Q \cdot (\frac{4}{3}\pi (2r)^3 - \frac{4}{3}\pi r^3) \cdot \frac{1}{\frac{4}{3}\pi(2R)^3 - \frac{4}{3}\pi R^3}. $$ Substituting this into the electric field equation, we can simplify and obtain \(E(r)\).

Case 3: \(r>2R\) (Points outside the shell)

For points located outside the shell, we again consider the Gaussian surface in the form of a sphere with radius \(r\). In this case, the Gaussian surface encloses both the sphere and the shell. Since the sphere has a total charge of \(+Q\) and the shell has a total charge of \(-Q\), the total enclosed charge is zero. Therefore, the electric field outside the shell is zero: $$ E(r) = 0. $$ Putting all these cases together, we obtain the electric field as a function of \(R\): $$ E(r) = \begin{cases} \frac{1}{\varepsilon_0} \cdot \frac{Q}{4\pi r^2} \cdot \frac{r^3}{R^3}, & r2R \end{cases} $$

Key concepts

Gauss's Law

Understanding the behavior of electric fields often begins with a fundamental principle known as Gauss's Law. This law is a cornerstone of electromagnetism and states that the electric flux through a closed surface, known as a Gaussian surface, is directly proportional to the total charge enclosed by that surface. Mathematically, Gauss's Law is expressed as:
\[\Phi_E = \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}\]where \(\Phi_E\) is the electric flux, \(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is the differential area vector on the closed surface S, \(Q_{\text{enc}}\) is the enclosed charge, and \(\varepsilon_0\) is the permittivity of free space. Gauss's Law provides a powerful tool for calculating electric fields, especially when dealing with symmetrical charge distributions, such as those in spherical conductors.

Uniform Charge Distribution

When a charge is spread out evenly over a volume or a surface, we refer to this as a uniform charge distribution. In the context of electric field calculations, knowing that the charge is uniformly distributed simplifies the process significantly. A common example featured in physics problems is the uniform distribution of charge within a sphere. The charge per unit volume, \(\rho\), remains constant, and it allows us to find the enclosed charge within any Gaussian surface inside the sphere by using the equation:
\[Q(r) = \rho \cdot V = \rho \cdot \frac{4}{3}\pi r^3\]where \(V\) represents the volume of the Gaussian sphere of radius \(r\). Knowing how the charge is spread is pivotal in applying Gauss's Law to find the electric field at any point inside the charged object.

Spherical Conductors

Spherical conductors are often used in physics problems to illustrate the principles of electrostatics due to their symmetry. In a conductor, charges are free to move until they reach a state where the electric field inside the conductor is zero. This results in the charges residing on the surface of the conductor. When dealing with hollow spherical conductors or spheres with uniformly distributed charge, the symmetry allows us to make certain assumptions. For instance, if we're measuring the electric field inside a hollow conducting sphere, it is zero everywhere inside the sphere since charges only reside on the exterior surface.

Electric Field Calculation

The electric field at a point in space is defined as the force per unit charge experienced by a small positive test charge placed at that point. For spherical objects with uniform charge distributions or conducting shells, the electric field's magnitude can be affected by the position relative to the charged object. As seen in the solved examples, to calculate the electric field within a uniformly charged sphere, we apply Gauss's Law at different ranges:

• Within the sphere, \(r < R\), where \(R\) is the radius of the sphere.
• Between the sphere and outer shell, \(R < r < 2R\).
• Outside the shell, where \(r > 2R\), leading to a zero electric field due to the cancellation of charges.

These calculations hinge on thoughtful choice of Gaussian surfaces and recognizing the symmetry of the charge distribution.

Gaussian Surface

A Gaussian surface is an imaginary closed surface used in conjunction with Gauss's Law to simplify the calculation of electric fields. The choice of Gaussian surface is key to effectively using Gauss's Law; the symmetry of the surface should match that of the charge distribution. For spherical charge distributions, a spherical Gaussian surface is chosen because the electric field at any point on this surface has the same magnitude and is directed radially outward. Choosing a spherical surface revolves around the perk that the dot product, \(\mathbf{E} \cdot d\mathbf{A}\), simplifies to \(E \cdot dA\), because the electric field vector and the area vector are parallel. This matching symmetry allows us to factor the electric field out of the integral, making the law highly convenient in the analysis of spherical charge distributions.