Question

A sphere centered at the origin has a volume charge distribution of \(120 . \mathrm{nC} / \mathrm{cm}^{3}\) and a radius of \(12.0 \mathrm{~cm} .\) The sphere is centered inside a conducting spherical shell with an inner radius of \(30.0 \mathrm{~cm}\) and an outer radius of \(50.0 \mathrm{~cm} .\) The charge on the spherical shell is \(-2.00 \mathrm{mC}\). What are the magnitude and the direction of the electric field at each of the following distances from the origin? a) at \(r=10.0 \mathrm{~cm}\) c) at \(r=40.0 \mathrm{~cm}\) b) at \(r=20.0 \mathrm{~cm}\) d) at \(r=80.0 \mathrm{~cm}\)

Short answer

Answer: The direction of the electric field in each case is as follows: 1. Case (a) - Inside the volume charge distribution sphere (r=10.0 cm): The electric field points radially outward. 2. Case (b) - Inside the conducting shell (r=20.0 cm): The electric field points radially outward. 3. Case (c) - Within the conducting shell itself (r=40.0 cm): There is no electric field. 4. Case (d) - Outside the conducting shell (r=80.0 cm): The electric field points radially inward since the conducting shell has a negative charge.

Step-by-step solution

Calculate the enclosed charge at r=10.0 cm

The volume of the sphere with radius r = 10.0 cm is: \(V_{sphere} = \frac{4}{3} \pi (10.0)^3\) Charge density is given as: \(ρ = 120 \times 10^{-9} C/cm^3 = 120 \times 10^{-3} C/m^3\) So, the enclosed charge is: \(q_{enclosed} = ρ \times V_{sphere}\)

Apply Gauss's Law

The Gaussian surface in this case is a sphere with a radius of 10.0 cm. The electric field is spherically symmetric, so Gauss's Law is applied: \(E \times 4 \pi (10.0)^2 = \frac{q_{enclosed}}{\epsilon_0}\) Solve for the electric field E. Electric field inside the conducting shell - Case b: r=20.0 cm

Find the total charge enclosed

Inside the conducting shell, the total charge enclosed is equal to the charge of the volume charge distribution sphere: \(q_{total} = \frac{4}{3} \pi (12.0)^3 × ρ\)

Apply Gauss's Law

Again, the Gaussian surface is a sphere with radius r=20.0 cm. The electric field is spherically symmetric, so Gauss's Law is applied: \(E \times 4 \pi (20.0)^2 = \frac{q_{total}}{\epsilon_0}\) Solve for the electric field E. Electric field within the conducting shell material - Case c: r=40.0 cm

Electric field inside a conductor

The electric field within a conducting material is zero due to the redistribution of charges that occurs. Therefore, in this case, the electric field at r=40.0 cm is: \(E = 0\) Electric field outside the conducting shell - Case d: r=80.0 cm

Calculate the enclosed charge

In this case, the total charge enclosed is the sum of the charge on the spherical shell and the charge of the volume charge distribution sphere: \(q_{enclosed} = q_{total} + q_{shell}\) where \(q_{shell} = -2.0 \times 10^{-3}C\)

Apply Gauss's Law

The Gaussian surface is a sphere with a radius of r=80.0 cm. The electric field is spherically symmetric, so Gauss's Law is applied: \(E \times 4 \pi (80.0)^2 = \frac{q_{enclosed}}{\epsilon_0}\) Solve for the electric field E. With the calculated electric field magnitude for each given distance, the directions can be determined using the following logic: 1. Case (a): Pointing radially outward 2. Case (b): Pointing radially outward 3. Case (c): No electric field 4. Case (d): Pointing radially inward (because the conducting shell has a negative charge)

Key concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that relates the distribution of electric charge to the resulting electric field. The law can be mathematically expressed using the equation:
\[\begin{equation}\oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{q_{\text{enc}}}{\varepsilon_0}\end{equation}\]where \(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is a vector representing an infinitesimal area on a closed surface S, with its magnitude being the area of the infinitesimal piece and its direction being perpendicular to the surface, \(q_{\text{enc}}\) is the charge enclosed by the surface S, and \(\varepsilon_0\) is the vacuum permittivity.
Gauss's Law is particularly useful for calculating the electric field in situations with high symmetry, such as spherical, cylindrical, or planar symmetry. In these cases, one can choose a Gaussian surface where the electric field is constant over the surface, making the calculations much simpler. For instance, when calculating the electric field around a spherical charge distribution or inside a spherical conductor, the Gaussian surface would also be a sphere concentric with the charge distribution, allowing the electric field to be easily determined.

Volume Charge Distribution

A volume charge distribution describes how electric charge is distributed throughout a three-dimensional region of space. It is often expressed in terms of charge density, \(\rho\), which is the amount of charge per unit volume. In mathematical terms, it is given as:
\[\begin{equation}\rho = \frac{dq}{dV}\end{equation}\]where \(dq\) is the infinitesimal amount of charge in an infinitesimal volume \(dV\). The total charge within a certain volume V is then calculated by integrating the charge density over that volume:
\[\begin{equation}q = \int_V \rho \ dV\end{equation}\]Understanding the volume charge distribution is crucial when applying Gauss's Law. The total enclosed charge in a given Gaussian surface directly influences the strength of the electric field. If the charge distribution has spherical symmetry, for example, this greatly simplifies the integration process, as was the case in the exercise above where we determined the charge within a spherical distribution.

Spherical Conductor

A spherical conductor is a conductive material shaped into a spherical form. When a charge is placed on such a conductor, it redistributes itself to the outer surface, leading to some unique properties. Notably, the electric field inside the conductor is zero due to the electrostatic shield effect. Even when external electric fields are present, the free charges in the conductor will move to counteract the external field, resulting in no net field inside.
Another important concept related to spherical conductors is that the electric field just outside the surface is perpendicular to the surface and can be calculated using Gauss's Law. The symmetry of a sphere allows us to conclude that the electric field is radially symmetric and can be treated as if all the charge were concentrated at the center of the sphere for points outside the conductor. This was used in the solution to determine the electric field outside the spherical conductor.

Electric Field Direction

The direction of the electric field is an integral part of understanding electromagnetic interactions. Electric fields originate from positive charges and terminate on negative charges. Following this principle, the field lines point away from positive charges and toward negative charges. In calculations using Gauss's Law for spherically symmetric situations, the direction is radially outward from the center of the sphere if the enclosed charge is positive, and radially inward if the charge is negative.
In a scenario with a conductive shell, the electric field inside the shell is zero and there are no field lines. However, outside the shell, the field lines will follow the direction expected from the net charge on the shell and any enclosed charges. As seen in the exercise, the direction of the electric field changed from pointing radially outward when only the positive volume charge distribution was considered, to radially inward when the negative charge on the outer conducting shell was also included.