Question

Free energy of charging a sphere. A sphere with radius \(a\) in a medium with dielectric constant \(D\) is uniformly filled with a charge of volume density \(\rho\). Derive the electrical free energy of this sphere in two different ways: (a) Derive the potential field and then charge the sphere from charge density 0 to \(\rho\). (b) Add fully charged shells of thickness \(d r\) from \(r=0\) to \(r=a\).

Short answer

The electrical free energy is \ \frac{2 \pi a^5 \rho^2}{15 \, D \, \epsilon_0} \

Step-by-step solution

Define Given Parameters

Given a sphere with radius \(a\) and dielectric constant \(D\), uniformly filled with charge density \ \rho \.

Step 2a: Derive the Potential Field (Method A)

Use Gauss's law for a spherical charge distribution. Inside the sphere, the electric field \(E(r)\) at a distance \(r\) from the center is given by: \[ \int_{0}^{r} D \, E(r') \, dA = \frac{q_{enc}}{\epsilon_0} \Rightarrow D \, E(r) \, 4 \pi r^2 = \frac{\rho \, \frac{4}{3} \pi r^3}{\epsilon_0} \ E(r) = \frac{\rho \, r}{3 \, D \, \epsilon_0} \] The potential \(V(r)\) is then found by integrating the electric field: \[ V(r) = -\int_0^r E(r') \, dr' = -\int_0^r \frac{\rho \, r'}{3 \, D \, \epsilon_0} \, dr' = -\frac{\rho}{6 \, D \, \epsilon_0} \int_0^r r' \, dr' = -\frac{\rho \, r^2}{6 \, D \, \epsilon_0} \] The potential at the surface of the sphere (r = a) is: \[ V(a) = -\frac{\rho \, a^2}{6 \, D \, \epsilon_0} \]

Step 3a: Calculate Free Energy by Charging

The free energy is given by the work done in assembling the charge. For continuous charge distribution: \[ W = \int_0^Q V(\rho') \, dQ = \int_0^\rho V(\rho') \, \frac{dQ}{d\rho} \, d\rho \] Substituting for \( V(\rho') \) and solving, we get: \[ W = \int_0^\rho -\frac{\rho' a^2}{6 \, D \, \epsilon_0} \cdot 4 \pi a^3 \, \rho' \, d\rho' = \frac{2 \pi a^5}{9 \, D \, \epsilon_0} \rho^2 \]

Step 2b: Fully Charged Shells (Method B)

Consider adding thin spherical shells of charge density from \(r = 0\) to \(r = a \). The electric field inside a shell can be calculated similarly: \[ E(r) = \frac{q_{enc}}{D \, \epsilon_0 \, 4 \pi r^2} \] The infinitesimal work to add a shell of thickness \(dr\) is: \[ dW = V(r) \, dQ = \left(\frac{\rho \, r^2}{6 \, D \, \epsilon_0} \right) \left( 4 \pi r^2 \, \rho \, dr \right) = \frac{2 \pi \rho^2 r^4}{3 \, D \, \epsilon_0} \, dr \]

Step 3b: Integrate to Find Total Work

Integrate from \(r = 0\) to 𝑎: \[ W = \int_0^a \frac{2 \pi \rho^2 r^4}{3 \, D \, \epsilon_0} \, dr = \frac{2 \pi \rho^2}{3 \, D \, \epsilon_0} \int_0^a r^4 \, dr = \frac{2 \pi \rho^2}{3 \, D \, \epsilon_0} \cdot \frac{a^5}{5} = \frac{2 \pi a^5}{15 \, D \, \epsilon_0} \rho^2 \]

Key concepts

electrostatics

Electrostatics deals with the study of electric charges at rest. It includes concepts like charge distribution, electric fields, and electric potential. In the context of our problem, electrostatics helps us understand how the charge density \(\rho\) is distributed uniformly within a sphere and influences the surrounding space. Analyzing the potential field and calculating the work needed to assemble the charge are key aspects relevant to electrostatics.

dielectric constant

The dielectric constant (D) of a medium indicates how well the medium can store electrical energy. It impacts the relationship between the electric field and the electric flux density. In our exercise, the dielectric constant of the medium surrounding the sphere plays a crucial role in determining the potential field and calculating the energy required for charging the sphere. A higher dielectric constant reduces the electric field strength for a given amount of charge.

Gauss's law

Gauss's law is one of Maxwell's four fundamental equations in electromagnetism. It states that the total electric flux through a closed surface is proportional to the charge enclosed by that surface: \[ \oint_{\text{surface}} \mathbf{D} \cdot d\mathbf{A} = Q_{\text{enc}} \] For spherical symmetry, this simplifies calculations. We used Gauss's law to determine the electric field inside the sphere: \[ E(r) = \frac{\rho \, r}{3 \, D \, \epsilon_0} \] This relationship helps us find the potential field and eventually derive the free energy of charging.

electric field

The electric field (E) describes the force per unit charge exerted on a test charge in the sphere's vicinity. For a sphere uniformly filled with charge density \( \rho \), we used Gauss's law to determine the field inside: \[ E(r) = \frac{\rho \, r}{3 \, D \, \epsilon_0} \] This expression tells us how the electric field varies within the sphere. Understanding the electric field is essential for calculating the potential field and work done during the charging process.

potential energy

Potential energy in electrostatics refers to the energy stored due to the positions of charges. For a sphere being charged, the potential energy corresponds to the work needed to assemble the charge. This involves integrating the potential over the charge density. Two approaches were used: (A) finding the potential field first and (B) considering thin shells of charge. Both methods aim to calculate the total work or free energy, showing consistency in the physical laws governing the system.

integration in physics

Integration is a crucial mathematical tool in physics for finding quantities like work and potential fields from known distributions, such as charge densities. In our solution, integration was used multiple times. First, to obtain the potential field: \[ V(r) = -\int_0^r E(r') \, dr' \] And then, to find the total work: \[ W = \int_0^a \frac{2 \pi \rho^2 r^4}{3 \, D \, \epsilon_0} \, dr = \frac{2 \pi a^5}{15 \, D \, \epsilon_0} \rho^2 \] Understanding how to set up and evaluate these integrals is essential for solving problems in electrostatics and calculating energy concepts.