Problem 55 - Fundamentals of Physics

Question

A charge distribution that is spherically symmetric but not uniform radially produces an electric field of magnitude \(E=K r^{4}\), directed radially outward from the center of the sphere. Here \(r\) is the radial distance from that center, and \(K\) is a constant. What is the volume density \(\rho\) of the charge distribution?

Step-by-Step Solution

Answer

The volume charge density is \( \rho(r) = 6\epsilon_0 K r^3 \).

Step 1: Applying Gauss's Law

To find the volume charge density \( \rho \), we use Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the vacuum permittivity \( \epsilon_0 \). Mathematically, it's expressed as \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\epsilon_0} \), where \( \mathbf{E} \) is the electric field. Here, the symmetry suggests using a spherical Gaussian surface with radius \( r \). Given \( E = K r^4 \), the electric field passing through the surface becomes \( 4\pi r^2 E = 4\pi r^2 (K r^4) \).

Step 2: Solving for Enclosed Charge

Now, calculate the charge \( Q_{enc} \) inside the sphere. Using the result from Step 1: \( 4\pi r^2 (K r^4) = \frac{Q_{enc}}{\epsilon_0} \) leads to \( Q_{enc} = 4\pi \epsilon_0 K r^6 \).

Step 3: Finding Volume Charge Density \(\rho\)

Volume charge density \( \rho \) is defined as the charge per unit volume. To find \( \rho \), differentiate the charge \( Q_{enc} = \int \rho \cdot dV \) with respect to volume \( V = \frac{4}{3} \pi r^3 \). Given \( Q_{enc} = 4\pi \epsilon_0 K r^6 \), differentiate with respect to volume: \( \rho(r) = \frac{dQ_{enc}/dr}{dV/dr} \). Since \( dV/dr = 4\pi r^2 \), then \( \rho(r) = \frac{d(4\pi \epsilon_0 K r^6)/dr}{4\pi r^2} = \frac{24\pi \epsilon_0 K r^5}{4\pi r^2} = 6\epsilon_0 K r^3 \).

Key Concepts

Spherical Symmetry

When we talk about spherical symmetry in physics, especially in the context of charge distributions, it means that the quantities we're looking at, like electric fields or charge density, depend only on the distance from a central point, not on the direction. It's like how concentric layers of an onion are arranged — every layer has the same shape around a central point.
Spherical symmetry simplifies calculations because you can envision this distribution as a series of concentric spheres.

• This symmetry dictates that the electric field will be directed radially outward or inward, depending on the charge.
• Gauss's Law takes advantage of this symmetry, allowing us to choose spherical Gaussian surfaces that enclose the charge symmetrically.

The beauty of spherical symmetry is that it reduces the complexity of three-dimensional problems to one dimension, centered around the distance from the center, noted as \( r \).

Electric Field

The electric field is a crucial concept in electromagnetism, representing the force per unit charge exerted on a positive test charge placed in the field. In our scenario, the electric field produced by the spherically symmetric charge distribution is directed radially outward.
The magnitude of this field at a distance \( r \) from the center is given by \( E = K r^4 \), where \( K \) is a constant. This equation tells us:

• The strength of the field grows significantly as \( r \) increases because of the \( r^4 \) dependence.
• The radial direction indicates that the force exerted by this field either pulls charges towards or pushes them away from the center, depending on the charge's sign.

Understanding the behavior of electric fields in such settings helps us apply Gauss's Law effectively to relate \( E \) to the charges producing it.

Charge Distribution

Charge distribution describes how charge is spread in a material or space. In a spherically symmetric scenario, the charge distribution depends solely on the radius \( r \). In our problem, although the charge distribution exhibits spherical symmetry, it varies with \( r \), which means it is not uniform radially.
This non-uniformity means the amount of charge within any given radius is different and must be calculated accordingly.
To handle this, Gauss's Law plays a vital role because it allows us to enclose any area with a Gaussian surface, letting us calculate the enclosed charge based on the electric field specifications. In this case, we derive that:

• The enclosed charge \( Q_{enc} \) initially relates to \( 4\pi \epsilon_0 K r^6 \), arising from integrating the electric field over a spherical surface.
• The evaluation of charge distribution leads to understanding how \( \rho(r) \), the charge density, varies with \( r \), interpreting the physical setup.

This connection between charge distributions and electric fields is key to understanding electromagnetism.

Volume Charge Density

Volume charge density \( \rho \) measures the amount of charge per unit volume at any point in space. It is a fundamental concept for understanding how charges are distributed within a volume.
In the problem, our objective is to ascertain \( \rho \) as a function of \( r \). We already have the enclosed charge \( Q_{enc} = 4\pi \epsilon_0 K r^6 \). To find \( \rho(r) \), consider the formula:

• \( \rho(r) = \frac{dQ_{enc}/dr}{dV/dr} \)

Here, \( dV/dr = 4\pi r^2 \) represents the rate of volume change with respect to the radius. With the derivative of \( Q_{enc} \) being \( 24\pi \epsilon_0 K r^5 \), we get:

• \( \rho(r) = \frac{24\pi \epsilon_0 K r^5}{4\pi r^2} = 6\epsilon_0 K r^3 \)

This equation reveals how charge density increases with \( r \), confirming the radial increase suggested by the electric field \( E = K r^4 \). Understanding \( \rho \) helps connect the electric field to the material's internal charge structure.