Problem 67 - University Physics with Modern Physics

Question

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. (\(\textit{Hint:}\) 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 \(dq\)? Then integrate to get the total energy.)

Step-by-Step Solution

Answer

The self-energy is \( \frac{3Q^2}{20\pi \varepsilon_0 R} \)."

Step 1: Calculate Charge Density

Since the charge is uniformly distributed throughout the volume of the sphere, we can calculate the charge density \( \rho \) using the formula for volume charge density. Given that the total charge is \( Q \) and the volume of the sphere is \( \frac{4}{3}\pi R^3 \), we have \( \rho = \frac{Q}{\frac{4}{3}\pi R^3} \).

Step 2: Consider the Electric Field Inside the Sphere

For a spherical charge distribution, the electric field at a distance \( r \) from the center inside the sphere is given by \( E(r) = \frac{1}{4\pi \varepsilon_0} \cdot \frac{1}{r^2} \cdot q(r) \), where \( q(r) \) is the charge enclosed within radius \( r \). Thus, \( q(r) = \rho \cdot \frac{4}{3} \pi r^3 = Q \cdot \left( \frac{r^3}{R^3} \right) \), so \( E(r) = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Qr}{R^3} \).

Step 3: Calculate Work Done to Add a Spherical Shell

The work done to bring a small charge \( dq \) to the sphere, at radius \( r \), is given by \( dW = E(r) \cdot dq \cdot dr \). Using the expression for \( E(r) \) and the charge of the shell \( dq = \rho \cdot 4\pi r^2 dr = Q \cdot \frac{r^2}{R^3} \cdot dr \), we get \( dW = \frac{Q^2}{4\pi \varepsilon_0 R^6} r^4 dr \).

Step 4: Integrate to Find Total Self-Energy

Integrate \( dW \) from \( r = 0 \) to \( r = R \) to find the total energy \( W \): \[ W = \int_0^R \frac{Q^2}{4\pi \varepsilon_0 R^6} r^4 dr = \frac{Q^2}{4\pi \varepsilon_0 R^6} \int_0^R r^4 dr. \] Evaluating the integral, \( \int_0^R r^4 dr = \frac{R^5}{5} \), gives \( W = \frac{Q^2}{4\pi \varepsilon_0 R^6} \cdot \frac{R^5}{5} = \frac{3Q^2}{20\pi \varepsilon_0 R} \).

Key Concepts

self-energy

The self-energy of a charge distribution is the total energy required to assemble a given amount of charge in a specific configuration. In the case of a uniformly charged solid sphere, self-energy is the work done to bring infinitesimal charges from infinity and assemble them into a solid sphere. This concept is crucial in electrostatics as it represents the energy stored within an electric field created by the charge distribution. To calculate this, we assume an incremental building-up of charge in spherical layers from the center to the surface, accounting for the work done at each stage.

uniform charge distribution

In the context of a sphere, a uniform charge distribution means the charge is spread evenly throughout the sphere's volume. This simplifies calculations because the charge density, defined as charge per unit volume, remains constant throughout the sphere. To express it mathematically, for a total charge \(Q\) in a sphere of radius \(R\), the volume is \(\frac{4}{3}\pi R^3\), and thus the charge density \(\rho\) is \(\rho = \frac{Q}{\frac{4}{3}\pi R^3}\). This uniform distribution allows us to predict the behavior of the electric field within the sphere, as well as calculate the energy needed to assemble the charge.

electric field inside sphere

The electric field inside a uniformly charged sphere varies with distance from the center. For any radius \(r\) within the sphere, the field is influenced only by the charge enclosed by the sphere of radius \(r\). According to Gauss's law, the field \(E(r)\) is proportional to \(r\) and is given by the formula:

• \(E(r) = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Qr}{R^3}\)

This linear relationship shows that the electric field strengthens as you move from the center (where \(r = 0\), hence \(E = 0\)) to the surface of the sphere. Understanding this field behavior is essential in calculating the work needed to bring additional charge to the sphere.

work done in assembling charge

To determine the work done in assembling the charge, we consider the energy required to bring in each infinitesimally small charge \(dq\) from infinity to the sphere's surface. For a spherical shell of thickness \(dr\) at radius \(r\), the work done \(dW\) is the product of the electric field \(E(r)\), charge \(dq\), and thickness \(dr\):

• \(dW = E(r) \cdot dq \cdot dr\)

Substituting in the charge for the shell, we derive \(dW = \frac{Q^2}{4\pi \varepsilon_0 R^6} r^4 dr\). Finally, integrating \(dW\) over the entire radius from 0 to \(R\) gives the total self-energy, \(W = \frac{3Q^2}{20\pi \varepsilon_0 R}\). This integration shows how the energy differs due to both the increasing electric field strength and increasing area as you move outward.