Question

Surface charge of uniform density \(\rho_{s}\) lies on a spherical shell of radius \(b\), centered at the origin in free space. ( \(a\) ) Find the absolute potential everywhere, with zero reference at infinity. (b) Find the stored energy in the sphere by considering the charge density and the potential in a two-dimensional version of Eq. (42). (c) Find the stored energy in the electric field and show that the results of parts \((b)\) and \((c)\) are identical.

Short answer

#Answer# 1. For r < b, E_in = 0; For r > b, E_out(r) = (b^2 * ρ_s) / (r^2 * ε_0) * r̂. 2. For r < b, V_in(b) = (b * ρ_s) / ε_0; For r > b, V_out(r) = (b^2 * ρ_s) / (r * ε_0). 3. Stored energy using charge density and potential: U = (2 * π * b^3 * ρ_s^2) / ε_0. 4. Stored energy in electric field: U_E = (2 * π * b^3 * ρ_s^2) / ε_0. 5. U = U_E, hence the results are identical.

Step-by-step solution

1. Calculate the electric field

To find the electric field, we must use Gauss's Law, which states: \begin{align} \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} \end{align} Here, \(Q_{enc}\) is the enclosed charge and \(\epsilon_0\) is the vacuum permittivity. The electric field will be different inside and outside the sphere. For \(r < b\), no charge is enclosed \((Q_{enc} = 0)\), so the electric field \(\vec{E}_{in}\) is zero: \begin{align} \vec{E}_{in}(r) = 0 \end{align} For \(r > b\), the electric field \(\vec{E}_{out}\) can be calculated. All the charge is enclosed, hence \(Q_{enc} = 4\pi b^2 \rho_s\): \begin{align} \vec{E}_{out}(r) = \frac{4\pi b^2 \rho_s}{4\pi r^2 \epsilon_0}\hat{r} = \frac{b^2 \rho_s}{r^2\epsilon_0}\hat{r} \end{align}

2. Find the potential

We now integrate the electric field to get the potential V(r), since \(V(r) = -\int \vec{E} \cdot d\vec{r}\).For \(r < b\), as the electric field is zero: \begin{align} V_{in}(r) = V_{in}(b) = V_{in}(b) - V_{\infty} = - \int_b^{\infty} \vec{E}_{out}(r) \cdot d\vec{r} \end{align} For \(r > b\), we have to integrate the electric field outside the sphere: \begin{align} V_{out}(r) = V_{out}(r) - V_{\infty} = - \int_r^{\infty} \vec{E}_{out}(r) \cdot d\vec{r} \end{align} Integrating the electric field \(E_{out}(r)\), we get: \begin{align} V_{in}(b) = \frac{b \rho_s}{\epsilon_0} \\ V_{out}(r) = \frac{b^2 \rho_s}{r \epsilon_0} \end{align}

3. Calculate stored energy using charge density and potential

To find the stored energy in the sphere, we will use the equation: \begin{align} U = \frac{1}{2} \int \rho_s V dV = \frac{1}{2} \rho_s V_{in}(b) \times 4 \pi b^2 \end{align} Plug in the value of \(V_{in}(b)\) and \(\rho_s\), we get: \begin{align} U = \frac{1}{2} \times \frac{b \rho_s}{\epsilon_0} \times (\rho_s \times 4\pi b^2) \end{align} Solving, we find: \begin{align} U = \frac{2\pi b^3 \rho_s^2}{\epsilon_0} \end{align}

4. Calculate stored energy in electric field

To find the stored energy in the electric field, we will use the energy density \(u_E = \frac{1}{2} \epsilon_0 E^2\): \begin{align} U_E = \int \frac{1}{2} \epsilon_0 E_{out}^2 dV = \frac{1}{2} \epsilon_0 \int_b^{\infty} E_{out}^2 \cdot 4\pi r^2 dr \end{align} Plug in the value of \(E_{out}(r)\), we get: \begin{align} U_E = \frac{1}{2} \epsilon_0 \int_b^{\infty} (\frac{b^2 \rho_s^2}{r^4\epsilon_0^2}) \cdot 4\pi r^2 dr \end{align} Integrate and solve to find: \begin{align} U_E = \frac{2\pi b^3 \rho_s^2}{\epsilon_0} \end{align} Comparing the results from part (b) and part (c), we see that: \begin{align} U = U_E \end{align} Hence, the results from part (b) and part (c) are identical.

Key concepts

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that provides a method to calculate the electric field produced by a given distribution of electric charge. According to Gauss's Law, the electric flux through a closed surface, known as a Gaussian surface, is equal to the charge enclosed by the surface divided by the vacuum permittivity \(\epsilon_0\). This principle is mathematically expressed as: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} \. \]
When applied to a spherical shell with uniform surface charge \(\rho_s\), as in our exercise, Gauss's Law reveals that the electric field inside the shell \(r < b\) is zero because no charge is enclosed by the Gaussian surface that lies within the shell. Conversely, for points outside the shell \(r > b\), the electric field is determined by the total charge on the shell's surface. The simplicity of this law lies in its ability to treat complex charge distributions with relative ease, making it a cornerstone for solving electrostatic problems.

Electric Potential

Electric potential represents the work done by an external force to move a unit positive charge from a reference point to a specific point in the electric field without producing an acceleration. Essentially, it quantifies the potential energy per unit charge at a point in space. For a spherical shell, the electric potential at a distance r from the center is zero at infinity and is calculated by integrating the electric field \(\vec{E}\) from the point of interest to infinity.

Inside the shell \(r < b\), since the electric field is zero, the potential remains constant and is the same as at the surface of the shell. For points outside \(r > b\), the potential decreases as \(\frac{1}{r}\), calculated using the integration of the electric field as shown in the exercise solution. This relationship between the electric field and potential is pivotal in understanding electric fields in terms of energy rather than force.

Stored Energy in Electrostatic Field

The stored energy in an electrostatic field, also known as electric field energy, quantifies the work required to assemble a given charge distribution. For the spherical shell, the energy stored is found by integrating the product of the charge density \(\rho_s\) and the electric potential \(V\) over the volume of the sphere. The energy can be expressed as \[ U = \frac{1}{2} \int \rho_s V dV. \]
This equation signifies that the energy is concentrated in the field surrounding the charge distribution. It is this energy that would be released if the charged spherical shell were allowed to discharge. The method of calculating stored energy based on the electric field directly, using the energy density formula \(u_E = \frac{1}{2} \epsilon_0 E^2\), substantiates the same result and showcases the equivalence of the electrostatic field energy approach to the charge density approach.

Integration of Electric Field

Integration of the electric field to determine the potential or stored energy is a process that involves calculating the cumulative effect of the electric field over a path or volume. In terms of potential, it's the integral of the electric field \(\vec{E}\) along a path that leads to determining the difference in potential between two points.

When calculating the stored energy in an electric field, we integrate over the volume of space. For our spherical shell, we integrate the squared electric field outside the shell \(E_{out}\) over all space extending from the surface of the shell to infinity. By integrating \(E_{out}^2\) times \(4\pi r^2 dr\), we accumulate the contribution of each shell of space to the total energy stored in the electrostatic field. This illustrates how energy is distributed throughout the space surrounding a charge distribution and how integration serves as a vital tool in these calculations.