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 through- out 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 \(d r\) having charge \(d q ?\) Then integrate to get the total energy.)

Step-by-Step Solution

Answer

The self-energy of the sphere is \( U = \frac{9k_e Q^2}{5R} \).

Step 1: Calculate Volume Charge Density

The volume charge density \( \rho \) is given by the total charge \( Q \) divided by the volume of the sphere of radius \( R \). The volume of a sphere is \( \frac{4}{3} \pi R^3 \), so the density is \( \rho = \frac{Q}{\frac{4}{3}\pi R^3} = \frac{3Q}{4\pi R^3} \).

Step 2: Consider Charge in Sphere of Radius r

For a smaller sphere of radius \( r \leq R \), the charge \( q \) contained in it is given by the volume charge density times the volume of this smaller sphere: \( q = \rho \cdot \frac{4}{3}\pi r^3 = \frac{Q}{R^3} r^3 \).

Step 3: Calculate Potential Energy to Add Charge dq

Consider adding a thin spherical shell of charge \( dq \) at radius \( r \). To compute its energy \( dU \), use the formula for energy \( dU = V \cdot dq \), where \( V \) is the electric potential due to charge \( q \) already assembled. The potential \( V \) at \( r \) is \( V = \frac{k_e q}{r} \), so \( dU = \frac{k_e q}{r} dq \).

Step 4: Express dq in terms of dr

The charge of the spherical shell is \( dq = \rho \cdot 4\pi r^2 dr = \frac{3Q}{4\pi R^3} 4\pi r^2 dr = \frac{3Q}{R^3} r^2 dr \). Substituting into our expression for \( dU \), we get \( dU = \frac{k_e}{r} \frac{Q}{R^3} r^3 \cdot \frac{3Q}{R^3} r^2 dr = \frac{9k_e Q^2}{R^6} r^4 dr \).

Step 5: Integrate to Find Total Self-Energy

Integrate \( dU \) from \( r = 0 \) to \( r = R \) to get the total self-energy: \[ U = \int_0^R \frac{9k_e Q^2}{R^6} r^4 dr = \frac{9k_e Q^2}{R^6} \left[ \frac{r^5}{5} \right]_0^R = \frac{9k_e Q^2}{R^6} \cdot \frac{R^5}{5} = \frac{9k_e Q^2}{5R} \].

Step 6: Final Expression for Self-Energy

The energy required to assemble the sphere of charge, known as the self-energy, is given by \( U = \frac{9k_e Q^2}{5R} \), where \( k_e = \frac{1}{4\pi\varepsilon_0} \) is the Coulomb's constant.

Key Concepts

Electrostatics

Electrostatics is the branch of physics that deals with the study of stationary electric charges or charges at rest. A primary focus of electrostatics is understanding how charges interact and influence each other when no movement occurs. In the context of the problem with the sphere, electrostatics principles help us understand how a charge is distributed within the sphere and how that affects the overall energy needed to assemble such a charge distribution.
One key aspect of electrostatics is the concept of field lines, which visualize how charges interact within a space. For instance:

• Field lines start from positive charges and end on negative ones.
• The density of these lines indicates the strength of the electric field in an area.

A good grasp of electrostatics ensures you can predict how charges influence each other over distance, which directly ties into concepts like electric potential and Coulomb's law.

Electric Potential

Electric potential, also known as potential difference, is a fundamental concept that describes the work needed to move a unit positive charge from one point to another within an electric field. When dealing with uniform charge distributions, like our charged sphere, understanding electric potential is vital for calculating self-energy, which is the energy required to assemble the charge configuration.
In this scenario:

• The potential at a particular point is determined by the charges present and their configuration.
• For instance, within the sphere, the potential at a radius \( r \) is used to calculate the work or energy needed to add a small charge \( dq \) at that radius.

The equation for electric potential from a point charge \( q \) at distance \( r \) is \( V = \frac{k_e q}{r} \), where \( k_e \) is Coulomb's constant. By integrating this potential over the volume, you derive the total energy required.

Volume Charge Density

Volume charge density, denoted as \( \rho \), measures how much electric charge is distributed within a given volume. It's a crucial term when dealing with continuous charge distributions, as seen with the sphere of charge in our exercise.
In our example:

• Volume charge density is calculated as the total charge \( Q \) divided by the sphere's volume, \( \frac{4}{3} \pi R^3 \).
• This gives us \( \rho = \frac{3Q}{4\pi R^3} \), which helps us determine how much charge there is in any part of the sphere, such as a smaller spherical volume of radius \( r \).

Understanding volume charge density allows us to calculate how charges are distributed and thus predict how they contribute to the overall electrostatic energy of the system.

Coulomb's Law

Coulomb’s Law is fundamental in electrostatics, providing a quantitative measure of the force between two point charges. It states that the force \( F \) between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, this force is given by:\[ F = k_e \frac{q_1 q_2}{r^2} \]where \( k_e \) is Coulomb's constant and \( r \) is the distance between the charges.
In the process of calculating the self-energy of the charged sphere:

• Coulomb's law helps determine the interaction between infinitesimal charges as they are brought together.
• It plays a key role in establishing the electric potential, \( V \), which is then used to calculate the energy required to assemble the charge distribution.

By understanding Coulomb's Law, you can determine the energetic interactions within the charged sphere, enabling a comprehensive analysis of the self-energy involved.