Question

A solid conducting sphere of radius \(r_{1}\) has a total charge of \(+3 Q\). It is placed inside (and concentric with) a conducting spherical shell of inner radius \(r_{2}\) and outer radius \(r_{3}\). Find the electric field in these regions: \(rr_{3}\)

Short answer

Answer: The magnitudes of the electric fields in the given regions are: 1. Inside the solid conducting sphere (r < r1): E = 0 2. Between the solid conducting sphere and the conducting spherical shell (r1 < r < r2): E(r) = (3Q) / (4πε₀r²) 3. Inside the conducting spherical shell (r2 < r < r3): E = 0 4. Outside the conducting spherical shell (r > r3): E(r) = (3Q) / (4πε₀r²)

Step-by-step solution

Inside the solid conducting sphere (\(r

Here, the electric field inside the solid conducting sphere \(r<r_1\) is equal to zero. This is because electric field inside a conductor is always zero in the electrostatic equilibrium state. Therefore, for \(r<r_1\), we get \(\textbf{E}=0\).

Between the solid conducting sphere and the conducting spherical shell (\(r_1

In this region, we can apply Gauss's law with a Gaussian surface, which would be a spherical surface concentric with the solid conducting sphere and the spherical shell. The Gaussian surface will have a radius of \(r\) such that \(r_1<r<r_2\). Since the electric field is radially symmetric, its magnitude will be constant at the Gaussian surface, and the electric flux through the surface becomes: \(\oint {\textbf{E}\cdot\textbf{dA}} = E(r) \cdot 4\pi r^2\) The enclosed charge in this region is equal to the charge on the solid conducting sphere, \(+3Q\). By Gauss's law, we get: \(\oint{\textbf{E}\cdot \textbf{dA}} = \frac{Q_\text{enc}}{\epsilon_0} \Rightarrow E(r) \cdot 4\pi r^2 = \frac{3Q}{\epsilon_0}\) So, the electric field for \(r_1<r<r_2\) is given by: \(E(r) = \frac{3Q}{4\pi \epsilon_0 r^2}\)

Inside the conducting spherical shell (\(r_2

Similarly to step 1, the electric field inside the conducting spherical shell will be zero, since it is a conductor. Therefore, for \(r_2<r<r_3\), we get \(\textbf{E} = 0\).

Outside the conducting spherical shell (\(r>r_3\))

Finally, in this region, we apply Gauss's law again with a Gaussian surface of radius \(r\) such that \(r>r_3\). The electric field is radially symmetric, and the electric flux through the Gaussian surface becomes: \(\oint {\textbf{E}\cdot\textbf{dA}} = E(r) \cdot 4\pi r^2\) Since the conducting sphere has a charge of \(+3Q\) and the conducting shell has an effect on the electric field outside it as if all of its charges were concentrated at its center, the total enclosed charge in this region is \(+3Q\). By Gauss's law, we get: \(\oint{\textbf{E}\cdot \textbf{dA}} = \frac{Q_\text{enc}}{\epsilon_0} \Rightarrow E(r) \cdot 4\pi r^2 = \frac{3Q}{\epsilon_0}\) So, the electric field for \(r>r_3\) is given by: \(E(r) = \frac{3Q}{4\pi \epsilon_0 r^2}\) In conclusion, the electric field in the given regions is as follows: 1. \(rr_3\): \(E(r) = \frac{3Q}{4\pi \epsilon_0 r^2}\)

Key concepts

Gauss's Law

Gauss's Law is a fundamental principle describing the relation between an electric charge and the resultant electric field. At the heart of this law is the concept that the electric flux through a closed surface, known as a Gaussian surface, is directly proportional to the enclosed charge. This law is mathematically represented as \( \oint \textbf{E}\cdot\textbf{dA} = \frac{Q_{\text{enc}}}{\epsilon_0} \) where \( \textbf{E} \) is the electric field, \( \textbf{dA} \) is the differential area vector on the Gaussian surface, \( Q_{\text{enc}} \) is the total charge enclosed, and \( \epsilon_0 \) is the vacuum permittivity. This equation demonstrates that the total flux out of the surface depends only on the charge within, not on the shape or size of the surface.

In the exercise concerning conducting spheres, Gauss's Law plays a critical role. It enables us to calculate the electric field in different regions by considering symmetrical Gaussian surfaces around the charge distribution. This leads to a simplification of complex problems into solvable equations, demonstrating the utility and power of Gauss's Law in electrostatics.

Electrostatic Equilibrium

Electrostatic equilibrium refers to a state where no net motion of charge occurs within a conductor. In this state, the electric field inside the conductor is zero. This is because the free charges inside a conductor redistribute themselves in response to an electric field until they no longer experience any force and the field inside becomes null. Thus, on a conducting surface, charges reside only on the exterior and establish an electric field in the space outside the conductor.

In our exercise with a solid conducting sphere and a conducting shell, the concept of electrostatic equilibrium explains why there is no electric field within the body of the conductors (\(r

Electric Flux

Electric flux is a measure of the electric field passing through a given area. It is conceptually represented as the number of electric field lines penetrating a surface. Mathematically, it is defined as the dot product of the electric field \(\textbf{E}\) and the area vector \(\textbf{A}\), which is \(\textbf{E}\cdot\textbf{A}\). The dot product accounts for the angle between the field lines and the normal to the surface, ensuring that only the perpendicular component of the field contributes to the flux.

In our scenario with the conducting spheres, we calculated electric flux to apply Gauss's Law. The surface area of our Gaussian surface is \(4\pi r^2\), and for a symmetrical charge distribution, the electric flux simplifies to \(E(r)\cdot 4\pi r^2\), making it possible to solve for the electric field \(E(r)\) as seen in the solution steps.

Spherical Symmetry

Spherical symmetry plays a pivotal role in simplifying the computation of electric fields around spherical charge distributions. When there is spherical symmetry, the charge distribution is uniform across any sphere centered at the center of symmetry. This symmetry implies that the electric field must also be radial and uniform at any distance from the center, leading to a simplification in the calculation of the electric field using Gauss's Law.

In the case of the conducting spheres, such symmetry allows us to assume that the electric field only has a radial component and is the same at all points equidistant from the center—in other words, the same on the surface of a sphere at any given radius. This significantly eases the analysis and calculation of electric fields, as shown in the establishment of the electric field in various regions around the conducting spheres in the exercise.

Electric Flux

Electric flux is a measure of the electric field passing through a given area. It is conceptually represented as the number of electric field lines penetrating a surface. Mathematically, it is defined as the dot product of the electric field \(\textbf{E}\) and the area vector \(\textbf{A}\), which is \(\textbf{E}\cdot\textbf{A}\). The dot product accounts for the angle between the field lines and the normal to the surface, ensuring that only the perpendicular component of the field contributes to the flux.

In our scenario with the conducting spheres, we calculated electric flux to apply Gauss's Law. The surface area of our Gaussian surface is \(4\pi r^2\), and for a symmetrical charge distribution, the electric flux simplifies to \(E(r)\cdot 4\pi r^2\), making it possible to solve for the electric field \(E(r)\) as seen in the solution steps.

Spherical Symmetry

Spherical symmetry plays a pivotal role in simplifying the computation of electric fields around spherical charge distributions. When there is spherical symmetry, the charge distribution is uniform across any sphere centered at the center of symmetry. This symmetry implies that the electric field must also be radial and uniform at any distance from the center, leading to a simplification in the calculation of the electric field using Gauss's Law.

In the case of the conducting spheres, such symmetry allows us to assume that the electric field only has a radial component and is the same at all points equidistant from the center—in other words, the same on the surface of a sphere at any given radius. This significantly eases the analysis and calculation of electric fields, as shown in the establishment of the electric field in various regions around the conducting spheres in the exercise.