Problem 70 - Fundamentals of Physics

Question

Charge of uniform volume density \(\rho=3.2 \mu \mathrm{C} / \mathrm{m}^{3}\) fills a nonconducting solid sphere of radius \(5.0 \mathrm{~cm} .\) What is the magnitude of the electric field (a) \(3.5\) \(\mathrm{cm}\) and (b) \(8.0 \mathrm{~cm}\) from the sphere's center?

Step-by-Step Solution

Answer

(a) Inside: \(4.48 \times 10^{4} \mathrm{N/C}\); (b) Outside: \(2.4 \times 10^{4} \mathrm{N/C}\).

Step 1: Understand the Problem

We need to find the electric field at two different points: one inside and one outside a solid nonconducting sphere with uniform charge density \( \rho = 3.2 \, \mu \mathrm{C}/\mathrm{m}^3 \). The sphere has a radius \( R = 5.0 \, \mathrm{cm} \). We will use Gauss's Law to find the electric field at \( r = 3.5 \, \mathrm{cm} \) and \( r = 8.0 \, \mathrm{cm} \).

Step 2: Use Gauss's Law for Inside the Sphere

For a point inside the sphere (\( r = 3.5 \, \mathrm{cm} \)), we use Gauss's Law: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]The charge enclosed by the Gaussian surface inside the sphere is given by: \[ Q_{\text{enclosed}} = \rho \cdot \frac{4}{3} \pi r^3 \]Substitute \( r = 3.5 \, \mathrm{cm} = 0.035 \, \mathrm{m} \) and calculate \( Q_{\text{enclosed}} \).Then, solve for \( E \) as \[ E \cdot 4 \pi r^2 = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]

Step 3: Calculation for Inside the Sphere

First calculate the enclosed charge: \[ Q_{\text{enclosed}} = 3.2 \times 10^{-6} \, \mathrm{C/m^3} \cdot \frac{4}{3} \pi (0.035)^3 \approx 5.5 \times 10^{-9} \, \mathrm{C} \]Now substitute back to find \( E \): \[ E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q_{\text{enclosed}}}{r^2} \approx 4.48 \times 10^{4} \, \mathrm{N/C} \]

Step 4: Use Gauss's Law for Outside the Sphere

For a point outside the sphere (\( r = 8.0 \, \mathrm{cm} \)), we use Gauss's Law assuming all charge is enclosed:\[ E \cdot 4\pi r^2 = \frac{Q_{\text{total}}}{\varepsilon_0} \]Where \( Q_{\text{total}} \) is the charge of the whole sphere:\[ Q_{\text{total}} = \rho \cdot \frac{4}{3} \pi R^3 \]Substitute \( R = 5.0 \, \mathrm{cm} = 0.05 \, \mathrm{m} \) to find \( Q_{\text{total}} \).

Step 5: Calculation for Outside the Sphere

Calculate the total charge of the sphere:\[ Q_{\text{total}} = 3.2 \times 10^{-6} \, \mathrm{C/m^3} \cdot \frac{4}{3} \pi (0.05)^3 \approx 1.7 \times 10^{-8} \, \mathrm{C} \]Now solve for \( E \) at \( r = 0.08 \, \mathrm{m} \):\[ E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q_{\text{total}}}{r^2} \approx 2.4 \times 10^4 \, \mathrm{N/C} \]

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle that relates the electric flux through a closed surface to the charge enclosed by that surface. This law is incredibly useful for determining electric fields in symmetrical situations, such as with spherical, cylindrical, or planar charge distributions. The mathematical formulation of Gauss's Law is given by the equation: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]Here,

• \(\oint \vec{E} \cdot d\vec{A}\) denotes the electric flux through a closed surface
• \(Q_{\text{enclosed}}\) is the total electric charge enclosed within that surface
• \(\varepsilon_0\) is the permittivity of free space

To apply Gauss's Law effectively, selecting an appropriate Gaussian surface is key. It should respect the symmetry of the problem and simplify calculations. This often results in either no electric field component or a constant electric field component crossing the surface, making it easy to solve for the electric field.

Uniform Charge Density

Uniform charge density refers to a situation where charge is evenly distributed across a volume, area, or length. In this problem, the charge density is given by \(\rho = 3.2 \, \mu \mathrm{C}/\mathrm{m}^3\), which indicates that every cubic meter of the sphere contains \(3.2\) microcoulombs of charge. The concept of uniform charge density simplifies calculations because we assume that every part of the charged object contains the same amount of charge per given volume. To calculate the total charge in a volume, the formula is:\[ Q = \rho \cdot V \]For a sphere, the volume \(V\) can be calculated using the formula:\[ V = \frac{4}{3} \pi R^3 \]Where \(R\) is the radius of the sphere. Thus, by knowing \(\rho\), you can easily determine the total charge present in the sphere or any subsection of it.

Nonconducting Sphere

A nonconducting sphere is a type of material where electric charges do not move freely. In the context of this problem, the sphere uniformly distributes its fixed charge but does not allow charge flow due to its nonconducting nature. The electrostatic properties are dictated by the initial charge distribution rather than any subsequent current flow, differentiating it from conducting materials where charges can redistribute. Understanding the behavior of electric fields within nonconductors is simpler because the field only depends on the original configuration of the charges. Thus, internal electric fields can be directly calculated using the distribution of these fixed charges and Gauss's Law.

Enclosed Charge

Enclosed charge refers to the amount of charge within a particular Gaussian surface. For Gauss's Law calculations, it is crucial to accurately determine this value, as it directly affects the calculated electric field. When dealing with a uniform charge density, the enclosed charge can easily be found as:\[ Q_{\text{enclosed}} = \rho \cdot V_{\text{enclosed}} \]For example, inside a sphere with a radius \(r\), smaller than the full radius \(R\), you calculate the enclosed volume using:\[ V_{\text{enclosed}} = \frac{4}{3} \pi r^3 \]This approach allows you to find the charge contained within any arbitrary radius, which is essential for applying Gauss's Law correctly to find the electric field at specific points inside or outside the charged sphere.