Problem 21 - University Physics with Modern Physics

Question

The electric field at a distance of 0.145 m from the surface of a solid insulating sphere with radius 0.355 m is 1750 N/C. (a) Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? (b) Calculate the electric field inside the sphere at a distance of 0.200 m from the center.

Step-by-Step Solution

Answer

Charge density \( \rho \approx 1.17 \times 10^{-8} \, C/m^3 \). The internal electric field at 0.200 m is approximately 1350 N/C.

Step 1: Understand the Problem

We're given an electric field at a point outside a solid sphere and asked to find the charge density and the electric field at an interior point. The sphere is uniformly charged, so we'll need Gauss's Law for both parts.

Step 2: Apply Gauss's Law for External Electric Field

For the sphere, when the point is outside, the electric field is given by the formula \( E = \frac{kQ}{r^2} \), where \( r \) is the distance from the center of the sphere, and \( Q \) is the total charge. Here, \( r = 0.355 + 0.145 = 0.500 \) m.

Step 3: Solve for Total Charge Q

Using \( E = 1750 \, \text{N/C} \), substitute in the formula: \( 1750 = \frac{kQ}{(0.500)^2} \). Solve for \( Q \): \( Q = \frac{1750 \times 0.500^2}{k} \). Where \( k = 8.99 \times 10^9 \, \text{N m}^2/ ext{C}^2 \).

Step 4: Calculate Total Charge Q

Substituting the values, \( Q = \frac{1750 \times 0.25}{8.99 \times 10^9} \approx 4.86 \times 10^{-9} \, C \).

Step 5: Find Volume of the Sphere

The volume of the sphere is given by \( V = \frac{4}{3} \pi r^3 \), where \( r = 0.355 \) m. Calculate this to find the volume.

Step 6: Compute the Charge Density \( \rho \)

Charge density \( \rho \) is \( \frac{Q}{V} \). Compute this using \( Q \) from Step 4 and \( V \) from Step 5.

Step 7: Calculate Electric Field Inside the Sphere

For points inside a uniformly charged sphere, \( E = \frac{kQr}{R^3} \), where \( r \) is the distance from the center and \( R \) is the sphere's radius. Here, \( r = 0.200 \) m and \( R = 0.355 \) m.

Step 8: Substitute Values to Find Internal Field

Use the internal field formula with \( Q = 4.86 \times 10^{-9} \, C \) and substitute \( r = 0.200 \) m and \( R \) into the equation to find the electric field.

Key Concepts

Electric Field Calculations

Electric field calculations can be a challenging task, but they become more manageable with the right understanding. When dealing with a solid insulating sphere with a uniform charge distribution, such calculations are essential.

To find the electric field at a point outside the sphere, we use Gauss's Law. This law relates the electric field to the total charge enclosed by a closed surface. For a point outside a sphere, it is as if all the charge is concentrated at the center. The formula used here is:

• \( E = \frac{kQ}{r^2} \)
• \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \)
• \( Q \) is the total charge
• \( r \) is the distance from the center of the sphere

For a point inside the sphere, the electric field depends on the distance from the center and is given by:

• \( E = \frac{kQr}{R^3} \)
• \( R \) is the sphere's radius

This depends on both the total charge and how far into the sphere you're measuring.

Charge Density

Charge density is a measure of how much charge is distributed over a certain space, and it is denoted by \( \rho \). Understanding this concept helps in solving problems related to electric fields.To find the charge density inside a sphere, the total charge \( Q \) calculated from the electric field is divided by the volume \( V \) of the sphere.

The formula for charge density is:

• \( \rho = \frac{Q}{V} \)

Where the volume \( V \) of a sphere is calculated using the formula:

• \( V = \frac{4}{3} \pi r^3 \)

This method allows us to distribute the total charge evenly over the volume of the sphere, giving a uniform charge density throughout the sphere.

Solid Insulating Sphere

A solid insulating sphere is a three-dimensional object that can resist the flow of electric charge. It holds a static distribution of charge because it cannot conduct—meaning charges don’t move around easily. This kind of sphere is often used in physics problems to simplify calculations. The insulating property ensures that the charge stays where it initially is, allowing us to use simplifications, such as assuming the charge is uniformly distributed.
When applying Gauss's Law, the insulating nature ensures a better-defined internal electric field, simplifying calculations both outside and inside the sphere.

Uniform Charge Distribution

Uniform charge distribution in a sphere means that charge is spread evenly throughout its volume. This concept greatly simplifies electric field calculations as it allows us to deploy simplified formulas. When charge is uniformly distributed:

• The external electric field is calculated as if all charge resides at the center.
• The internal field changes linearly with distance from the center.

This idealized setup makes it easier to apply mathematical models, ensuring predictable behaviors of fields both inside and outside the insulating sphere.