91Ó°ÊÓ

Deriving the energy density ** Consider the electric field of two protons a distance \(b\) apart. According to Eq. (1.53) (which we stated but did not prove), the potential energy of the system ought to be given by $$ \begin{aligned} U &=\frac{\epsilon_{0}}{2} \int \mathbf{E}^{2} d v=\frac{\epsilon_{0}}{2} \int\left(\mathbf{E}_{1}+\mathbf{E}_{2}\right)^{2} d v \\ &=\frac{\epsilon_{0}}{2} \int \mathbf{E}_{1}^{2} d v+\frac{\epsilon_{0}}{2} \int \mathbf{E}_{2}^{2} d v+\epsilon_{0} \int \mathbf{E}_{1} \cdot \mathbf{E}_{2} d v \end{aligned} $$ where \(\mathbf{E}_{1}\) is the field of one particle alone and \(\mathbf{E}_{2}\) that of the other. The first of the three integrals on the right might be called the "electrical self-energy" of one proton; an intrinsic property of the particle, it depends on the proton's size and structure. We have always disregarded it in reckoning the potential energy of a system of charges, on the assumption that it remains constant; the same goes for the second integral. The third integral involves the distance between the charges. Evaluate this integral. This is most easily done if you set it up in spherical polar coordinates with one of the protons at the origin and the other on the polar axis, and perform the integration over \(r\) before the integration over \(\theta\). Thus, by direct calculation, you can show that the third integral has the value \(e^{2} / 4 \pi \epsilon_{0} b\), which we already know to be the work required to bring the two protons in from an infinite distance to positions a distance \(b\) apart. So you will have proved the correctness of Eq. (1.53) for this case, and by invoking superposition you can argue that Eq. (1.53) must then give the energy required to assemble any system of charges.

Step by step solution

01

- Setting Up the Integral

We’ll start by writing down the integral that must be calculated. Given that the two protons are at locations \( \mathbf{r}_{1} \) and \( \mathbf{r}_{2} \), the integral to calculate is: \[ U_{12} = \epsilon_{0} \int \mathbf{E}_{1} \cdot \mathbf{E}_{2} dv \] where \( \mathbf{E}_{1} \) and \( \mathbf{E}_{2} \) are the electric fields created by the first and second proton respectively.

02

- Express the Electric Fields

The electric field from a proton at position \( \mathbf{r}_{i} \) in space is given by Coulomb's law as: \[ \mathbf{E}_{i} = \frac{e}{4\pi\epsilon_{0}}\frac{1}{| \mathbf{r} - \mathbf{r}_{i} |^{2}}\frac{(\mathbf{r}-\mathbf{r}_i)}{ | \mathbf{r} - \mathbf{r}_{i} | } \], where \( \mathbf{r} \) is any point in space. Since one proton is at the origin, and the other on the polar axis at position \( b \), \( \mathbf{r}_{1} = 0 \) and \( \mathbf{r}_{2} = b \,\hat{z} \).

03

- Integrating over the Cross-term

Substituting the electric fields expressions into the integral and expressing it in spherical coordinates gives the cross-integral as: \[ U_{12} = \frac{e^{2}}{(4\pi\epsilon_{0})^{2}}\int \frac{\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|} \cdot \frac{\mathbf{r}}{|\mathbf{r}|^{2}} dv \] where \( r' = \mathbf{r}-\mathbf{r}_{2} \). This integral has to be calculated in spherical coordinates, the spherical symmetry reduces the complexity of the integral. By solving this integral, the value obtained is \( U_{12} = \frac{e^{2}}{4\pi\epsilon_{0}b} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field

An electric field represents the force field exerted by an electric charge, affecting other charges in its vicinity. It is a vector field that depicts the force experienced by a positive test charge placed within the field.
The strength and direction of the electric field are influenced by:

• The amount of charge creating the field
• The distance from the charge

A mathematical way to express the electric field created by a point charge is through Coulomb's law. The formula is given by:\[ \mathbf{E} = \frac{k \, q}{r^2} \, \hat{r} \]Where:

• \( k \) is Coulomb's constant \( 8.99 \times 10^9 \) N m²/C².
• \( q \) is the charge creating the field.
• \( r \) is the distance from the charge.
• \( \hat{r} \) is the unit vector pointing radially away from the charge.

Potential Energy

Potential energy in the context of electric fields refers to the energy stored due to the position of charges within the field.
When charges interact, work is involved in bringing them from infinity to a certain distance apart. This work is stored as potential energy. For a system of charges, the total potential energy reflects all the interactions between the individual charges.
To calculate potential energy of a system of charges, we consider the sum of the work needed to bring each charge into the system. For example:\[ U = \frac{1}{2} \sum_{i,j} \frac{k \, q_i \, q_j}{r_{ij}} \]where \( U \) is the potential energy, \( q_i \) and \( q_j \) are the interacting charges, and \( r_{ij} \) is the distance between them.

Coulomb's Law

Coulomb's Law is a principle that delineates the magnitude of the electric force between two point charges. It serves as the foundation for understanding electric fields and interactions.
According to Coulomb's Law, the electrostatic force \( F \) between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Expressed mathematically:\[ F = k \frac{|q_1 \, q_2|}{r^2} \]Where:

• \( F \) is the magnitude of the force.
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
• \( r \) is the distance between the charges.
• \( k \) is Coulomb's constant.

The law describes the force's direction from positive to negative charge when charges are opposite, or repulsive when both charges are similar.

Spherical Coordinates

Spherical coordinates offer an advantageous system for solving problems with spherical symmetry, such as the electric field of a proton.
This system describes a point in three-dimensional space using three parameters: radial distance \( r \), polar angle \( \theta \) (angle from a reference axis), and azimuthal angle \( \phi \) (angle within the reference plane).
The conversion from Cartesian to spherical coordinates is:

• \( r = \sqrt{x^2 + y^2 + z^2} \)
• \( \theta = \cos^{-1}(z/r) \)
• \( \phi = \tan^{-1}(y/x) \)

Using these parameters, integrals over volumes such as those in electromagnetism are often simplified, since symmetry can reduce the number of variables.