Problem 24 - University Physics with Modern Physics

Question

22.24. A point charge of \(-2.00 \mu \mathrm{C}\) is located in the center of a spherical cavity of radius 6.50 \(\mathrm{cm}\) inside an insulating charged solid. The charge density in the solid is \(\rho=7.35 \times 10^{-4} \mathrm{C} / \mathrm{m}^{3} .\) Calculate the electric field inside the solid at a distance of 9.50 \(\mathrm{cm}\) from the center of the cavity.

Step-by-Step Solution

Answer

The electric field at 9.50 cm is determined by combining the charge density effects and the point charge influence on the Gaussian surface.

Step 1: Understand the problem

We have a spherical cavity inside an insulating solid, with a point charge at its center. We want to find the electric field at a point inside the solid, specifically at a radius of 9.50 cm from the center of the cavity.

Step 2: Identify the charge distribution

The charge distribution is uniform throughout the solid, with the exception of the cavity, and has a density of \( \rho = 7.35 \times 10^{-4} \, \mathrm{C/m^3} \). The point charge at the center of the cavity is \( q = -2.00 \, \mu C = -2.00 \times 10^{-6} \, C \).

Step 3: Apply Gauss's Law

We can use Gauss's Law to find the electric field: \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \). The Gaussian surface is a sphere of radius 9.50 cm centered at the cavity center.

Step 4: Calculate the enclosed charge

Inside the Gaussian sphere (excluding the cavity), the charge from the solid is given by \( Q_{solid} = \rho \cdot V = \rho \cdot \frac{4}{3} \pi (0.095^3 - 0.065^3) \). Calculate this value.

Step 5: Account for the cavity's point charge

The total enclosed charge \( Q_{enc} \) includes the charge from the point charge at the center. Thus, \( Q_{enc} = Q_{solid} + (-2.00 \times 10^{-6} \mathrm{C}) \).

Step 6: Calculate the electric field

Knowing \( Q_{enc} \), compute the electric field using Gauss's Law: \( E = \frac{Q_{enc}}{4\pi\varepsilon_0 r^2} \), where \( r = 0.095 \mathrm{m} \).

Step 7: Final Calculation

Substitute the quantities into the formula. Use \( \varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C^2/N \cdot m^2} \) to find the electric field. Perform the calculation.

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle used to relate the electric flux flowing through a surface to the charge enclosed within that surface. It is expressed as: \[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \]where \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is the differential area vector, \( Q_{enc} \) is the total charge enclosed by the surface, and \( \varepsilon_0 \) is the permittivity of free space. This law applies best to symmetrical charge distributions.
Gauss's law is especially useful when dealing with symmetrical geometries like spheres, cylinders, and planes. It simplifies the calculation of electric fields by allowing us to consider only the charge enclosed by an imaginary Gaussian surface. In the case of a sphere, if a point charge is placed at the center, the law helps us determine the electric field surrounding it by considering a spherical Gaussian surface.
In our specific problem, we use Gauss's law to find the electric field at a distance from the center of a charged sphere with a cavity. This involves choosing the appropriate Gaussian surface at 9.50 cm from the center of the cavity.

Charge Distribution

Charge distribution refers to how electric charge is spread out in a given volume or over a surface. It can be uniform or non-uniform. In our exercise, we have a uniform charge distribution across the solid region of the sphere, except inside the cavity where the point charge is located.
Understanding this distribution is key to solving problems involving electric fields. We denote the charge density \( \rho \), which signifies the amount of charge per unit volume. In this scenario, the charge density is given as \( \rho = 7.35 \times 10^{-4} \, \mathrm{C/m^3} \).
For any point outside the cavity but inside the sphere, the electric field is affected by both the charge within that volume and the point charge at the center. The charge distribution affects how much total charge is enclosed by a Gaussian surface and thus influences the resulting electric field. By integrating the volume from the cavity's edge to the desired point, we calculate the charge enclosed by the chosen Gaussian surface.

Electric Charge

Electric charge is a fundamental property of matter that causes it to experience a force in an electromagnetic field. In our problem, the presence of a point charge inside the spherical cavity plays a crucial role in determining the electric field in the surrounding solid.
The point charge is

• negative, with a value of \( q = -2.00 \, \mu C = -2.00 \times 10^{-6} \, C \)

This negative charge creates an electric field that opposes the field created by the positive charge density of the surrounding solid.
Combining the effects of the negative point charge and the uniformly distributed positive charge density in the solid allows us to calculate the net electric field at a given point away from the cavity. These charges collectively determine the direction and magnitude of the electric field by being the source of the electric forces within the system.

Spherical Cavity

A spherical cavity in the context of our problem refers to a hollow space inside an insulated solid sphere that contains a point charge at its center. The cavity affects the distribution of electric field lines and the calculation of the electric field outside it.
Although the cavity itself doesn't have any electric charge from the solid, it houses a significant point charge, influencing the electric field within and beyond the cavity.
Here’s what you need to consider about spherical cavities:

• The cavity radius defines the region without the solid's charge.
• Electric field within the cavity is solely due to the point charge present.
• Effects on the solid's charge distribution are only outside the cavity.

These properties are critical when applying Gauss's Law. The law enables the calculation of the electric field in the solid region beyond the cavity by considering both the solid's charge and the point charge hidden within the cavity. Thus, the spherical cavity significantly determines where and how to apply the principles affecting the problem's solution.