Problem 65 - Fundamentals of Physics

Question

Charge \(Q\) is uniformly distributed in a sphere of radius \(R\). (a) What fraction of the charge is contained within the radius \(r=R / 2.00 ?\) (b) What is the ratio of the electric field magnitude at \(r=R / 2.00\) to that on the surface of the sphere?

Step-by-Step Solution

Answer

(a) The fraction of charge is \( \frac{1}{8} \). (b) The ratio of electric fields is \( \frac{1}{2} \).

Step 1: Understanding the Problem

We have a sphere with a charge \( Q \) uniformly distributed across its volume. We need to find the fraction of charge inside a smaller sphere of radius \( r = R/2 \), and the ratio of the electric field magnitude at \( r = R/2 \) compared to the electric field at the surface.

Step 2: Calculating Charge Density

Since the charge \( Q \) is uniformly distributed, the volume charge density \( \rho \) is \( \rho = \frac{Q}{\frac{4}{3}\pi R^3}\). This is because the total charge \( Q \) is spread over the entire volume of the sphere.

Step 3: Volume of Smaller Sphere

Calculate the volume of the sphere of radius \( r = R/2 \). The formula for the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \). So, \( V_{small} = \frac{4}{3}\pi (R/2)^3 = \frac{4}{3}\pi \frac{R^3}{8} = \frac{1}{6}\pi R^3 \).

Step 4: Charge in the Smaller Sphere

The charge inside the smaller sphere is \( Q_{inside} = \rho \times V_{small} = \left(\frac{Q}{\frac{4}{3}\pi R^3}\right) \times \frac{1}{6}\pi R^3 = \frac{Q}{8} \).

Step 5: Fraction of Charge Inside r = R/2

The fraction of the total charge inside the smaller sphere is \( \frac{Q_{inside}}{Q} = \frac{Q/8}{Q} = \frac{1}{8} \).

Step 6: Electric Field Inside a Uniformly Charged Sphere

Using Gauss's Law, the electric field at a distance \( r \leq R \) from the center in a uniformly charged sphere is \( E = \frac{1}{4\pi \varepsilon_0} \frac{Q_{inside}}{r^2} = \frac{1}{4\pi \varepsilon_0} \frac{Qr}{R^3} \), where \( Q_{inside} = \frac{Qr^3}{R^3} \).

Step 7: Electric Field at r = R/2

Substitute \( r = R/2 \) into the expression for \( E \). The electric field is \( E_{r=R/2} = \frac{1}{4\pi \varepsilon_0} \frac{Q(R/2)}{R^3} = \frac{1}{4\pi \varepsilon_0} \frac{QR}{2R^3} = \frac{Q}{8\pi \varepsilon_0 R^2} \).

Step 8: Electric Field on the Surface

The electric field at the surface \( r = R \) is \( E_{r=R} = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R^2} \).

Step 9: Ratio of Electric Fields

The ratio of the electric field magnitude at \( r = R/2 \) to that on the surface is \( \frac{E_{r=R/2}}{E_{r=R}} = \frac{\frac{Q}{8\pi \varepsilon_0 R^2}}{\frac{Q}{4\pi \varepsilon_0 R^2}} = \frac{1}{2} \).

Key Concepts

Electric Field

The electric field describes the space around a charged object where a force would be experienced by another charge. It is a vector field and is represented by the symbol \( E \). The concept of an electric field is foundational in understanding how charges interact without direct contact.
In the case of a uniformly charged sphere, Gauss's Law can be utilized to calculate the electric field at any point inside or outside the sphere. For a point inside the sphere, the electric field is not uniform and depends on the distance from the center. At a point at distance \( r \) within the sphere (where \( r < R \)), the electric field is given by:

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

Here, \( Q \) is the total charge, \( r \) is the distance from the center, \( R \) is the radius of the sphere, and \( \varepsilon_0 \) is the permittivity of free space. The calculation of the electric field involves understanding how the charge is distributed within a volume.

Charge Distribution

Charge distribution refers to how a charge is arranged in space. In a uniformly charged sphere, the charge is spread evenly throughout the volume, much like sugar spread evenly throughout a ball of dough.
This uniform spread means each small volume of the sphere has the same amount of charge per unit volume. This is quantified as the charge density, \( \rho \), which is calculated as:

• \( \rho = \frac{Q}{\frac{4}{3}\pi R^3} \)

Understanding charge distribution is crucial when applying Gauss's Law to determine electric fields. Inside a uniformly charged sphere, the charge affecting the electric field at any point is considered to be only that inside a sphere of radius \( r \), where \( r \) is the distance from the center of the sphere to the point of interest. By understanding the charge distribution, we can determine the enclosed charge at any radius to find the electric field there.

Sphere of Charge

A sphere of charge is a three-dimensional object where electric charge is distributed over a spherical volume. This concept is often explored in problems using Gauss's Law due to its symmetrical properties, which simplify calculations.
When dealing with a sphere of charge, Gauss's Law is an invaluable tool. It states that the electric flux through a closed surface is proportional to the enclosed charge. For a uniformly charged sphere, the law helps in calculating the electric field at any point by considering a spherical "Gaussian surface" within or around the sphere.
To solve problems involving a sphere of charge, it is essential to apply:

• The formula for the volume of a sphere: \( V = \frac{4}{3}\pi r^3 \)
• The concept of charge density and enclosed charge

Moreover, knowing that the electric field is radial and points outward helps in visualizing how charges affect their surroundings. This understanding is pivotal in solving exercises that require calculating fields at various points in and around the sphere.