Problem 71 - University Physics with Modern Physics

Question

Self-Energy of a Sphere of Charge. A solid sphere of radius \(R\) contains a total charge \(Q\) distributed uniformly throughout its volume. Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is called the "self-energy" of the charge distribution. (Hint: After you have assembled a charge \(q\) in a sphere of radius \(r,\) how much energy would it take to add a spherical shell of thickness dr having charge \(d q ?\) Then integrate to get the total energy.)

Step-by-Step Solution

Answer

The self-energy of the sphere is \( U = \frac{3kQ^2}{5R} \).

Step 1: Understand the Problem

We need to find the energy required to assemble a charge distribution by bringing infinitesimal charges to form a sphere of radius \( R \) from a distance.

Step 2: Calculate Charge Density

The total charge \( Q \) is uniformly distributed over the sphere. The charge density \( \rho \) can be calculated as \( \rho = \frac{Q}{\frac{4}{3}\pi R^3} \).

Step 3: Find Charge in Sphere Radius r

The charge \( q \) within a sphere of radius \( r \) is \( q = \frac{Q}{\frac{4}{3}\pi R^3} \cdot \frac{4}{3}\pi r^3 = Q\frac{r^3}{R^3} \).

Step 4: Energy to Add Infinitesimal Charge

The differential energy \( dU \) to add a shell of thickness \( dr \) with charge \( dq \) is given by the formula: \( dU = \frac{k q}{r} dq \). Here, \( dq = \rho \cdot 4\pi r^2 dr \).

Step 5: Substitute dq And Simplify

Substitute \( dq = \rho \cdot 4\pi r^2 dr \) into the energy expression: \( dU = \frac{k Q r^2}{R^3} \cdot \frac{4\pi r^2}{r} dr = \frac{4\pi k Q}{R^3} r^3 dr \). Simplify further: \( dU = \frac{4\pi k Q}{R^3} r^3 dr \).

Step 6: Integrate to Find Total Energy

Integrate over the entire volume to find the total self-energy: \( U = \int_0^R \frac{4\pi k Q}{R^3} r^3 dr \). Carrying out the integration, \( U = \frac{4\pi k Q}{R^3} \cdot \frac{r^4}{4} \Bigg|_0^R = \frac{4\pi k Q}{R^3} \cdot \frac{R^4}{4} = \frac{3kQ^2}{5R} \).

Step 7: Final Result

The self-energy of the charge distribution in the sphere is \( U = \frac{3kQ^2}{5R} \).

Key Concepts

Charge Distribution

In physics, understanding how charge is spread out in space is crucial. Charge distribution refers to how an electric charge is dispersed over a given space or object. For this exercise, the charge is distributed uniformly inside a solid sphere with a radius of \( R \). This means that every part of the volume within the sphere has the same amount of charge per unit volume.

To express this mathematically, we use charge density \( \rho \), defined as the total charge \( Q \) divided by the volume of the sphere, which is \( \frac{4}{3}\pi R^3 \). Therefore, the charge density is \( \rho = \frac{Q}{\frac{4}{3}\pi R^3} \). Uniform distribution implies that any smaller sphere of radius \( r \) within the original sphere will also have a proportionate charge, described as \( q = Q \frac{r^3}{R^3} \). Understanding these principles sets the stage for calculating other aspects like potential energy.

Electric Potential Energy

Electric potential energy is the energy stored due to the relationships between charged particles. For a sphere with charge distributed throughout, calculating the energy needed to assemble it is insightful. This energy, known as self-energy, requires integrating the electric potential contributions from infinitesimally small charge segments.

To find the self-energy, consider incrementally adding small shells of charge, \( dq \), to an existing sphere of radius \( r \). Each time a new shell is added, it turns out the energy required to place it can be given by \( dU = \frac{k q}{r} dq \), where \( k \) is the Coulomb's constant. This step combines understanding of basic physics with mathematical modeling, providing insight into how energy requirements change with increasing charge or size of sphere.

The key is recognizing the need to account for every small addition. Thus, by calculating \( dU \) using earlier determined charge density expressions, you take a significant leap towards determining the total self-energy of the charge within the sphere.

Integration in Physics

Integration is a powerful mathematical tool, essential for analyzing physical systems. In this context, integration is crucial to determine the total self-energy of a uniformly charged sphere. After breaking down the energy required to add infinitesimal charges in the previous step, integration accumulates these small energy changes over the entire volume of the sphere.

The integral \( U = \int_0^R \frac{4\pi k Q}{R^3} r^3 dr \) sums up the differential energy \( dU = \frac{4\pi k Q}{R^3} r^3 dr \) across the sphere from radius \( 0 \) to \( R \). This is a step-by-step addition of each small contribution from all shells extending from the center to the edge of the sphere.

After computing the integral, the outcome represents total potential energy stored due to the presence of the charge distribution. The result you achieve is \( U = \frac{3kQ^2}{5R} \), characterizing how integration translates a seemingly complex problem of continuous charge into finite energy solutions.