Problem 78 - Fundamentals of Physics

Question

A charge of \(6.00 \mathrm{pC}\) is spread uniformly throughout the volume of a sphere of radius \(r=4.00 \mathrm{~cm}\). What is the magnitude of the electric field at a radial distance of (a) \(6.00 \mathrm{~cm}\) and (b) \(3.00 \mathrm{~cm}\) ?

Step-by-Step Solution

Answer

(a) Electric field is approximately 29.9 N/C, (b) Electric field is approximately 14.9 N/C.

Step 1: Understanding the Problem

We are given a sphere with a charge of \(6.00 \text{ pC}\) uniformly distributed over a radius of \(4.00 \text{ cm}\). The problem asks us to find the electric field at two radial distances: outside the sphere at \(6.00 \text{ cm}\) and inside the sphere at \(3.00 \text{ cm}\).

Step 2: Convert Units

First, convert all units into SI units. Since 1 picoCoulomb (pC) is \(10^{-12}\) C, the charge is \(6.00 \times 10^{-12} \text{ C}\). The radius is \(4.00 \text{ cm} = 0.04 \text{ m}\), and distances are \(6.00 \text{ cm} = 0.06 \text{ m}\) and \(3.00 \text{ cm} = 0.03 \text{ m}\).

Step 3: Use Gauss's Law for (a)

For a radial distance \(r = 6.00 \text{ cm}\), which is outside the sphere, use Gauss's law. Gauss's law states that \( E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \), where \(E\) is the electric field, \(r = 0.06\text{ m}\), \(Q = 6.00 \times 10^{-12} \text{ C}\), and \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2)\).

Step 4: Calculate Electric Field (a)

Substitute into Gauss's law to find \(E\): \[ E = \frac{6.00 \times 10^{-12}}{4\pi (0.06)^2 \times 8.85 \times 10^{-12}} \]Calculate this to find \(E\).

Step 5: Use Gauss's Law for (b)

For a radial distance \(r = 3.00 \text{ cm}\), which is inside the sphere, Gauss's law requires the charge enclosed by the Gaussian surface to be considered. The charge density \(\rho\) is \(\frac{Q}{\frac{4}{3}\pi (0.04)^3}\). The charge enclosed for \(r = 0.03 \text{ m}\) is \(\rho \times \frac{4}{3}\pi (0.03)^3\).

Step 6: Calculate Electric Field (b)

Now use Gauss's law for the smaller sphere:\[ E\cdot 4\pi (0.03)^2 = \frac{\rho \times \frac{4}{3}\pi (0.03)^3}{\varepsilon_0} \]Simplify to find\[ E = \frac{\rho \cdot 0.03}{3\varepsilon_0} \]Calculate \(E\) using \(\rho = \frac{6.00 \times 10^{-12}}{\frac{4}{3}\pi (0.04)^3}\).

Key Concepts

Gauss's law

Gauss's law is a powerful tool used to understand electric fields. It relates the electric flux passing through a closed surface to the charge enclosed by that surface. Mathematically, Gauss’s law is stated as:

• \( E \cdot A = \frac{Q}{\varepsilon_0} \)

where \(E\) is the electric field, \(A\) is the area of a closed surface, \(Q\) is the total enclosed charge, and \(\varepsilon_0\) is the vacuum permittivity. By choosing a surface known as a "Gaussian surface," we can simplify calculations for symmetrical charge distributions, such as spheres. Gauss's law is particularly handy because it allows us to calculate electric fields without complex integrations. For the sphere in our exercise, we choose spherical surfaces both inside and outside the charged sphere to make use of symmetry.

Charge density

Charge density, often represented by \(\rho\), describes how charge is spaced throughout a given volume. For a uniform distribution of charge, like in the exercise, the charge density is defined as the total charge divided by the volume:

• \( \rho = \frac{Q}{V} \)
• For a sphere, \(V = \frac{4}{3} \pi r^3\).

In our example, the charge \(Q\) is \(6.00 \times 10^{-12} \text{ C}\), and the radius of the sphere is \(0.04 \text{ m}\). This leads to a charge density of:

• \( \rho = \frac{6.00 \times 10^{-12}}{\frac{4}{3} \pi (0.04)^3} \)

Understanding charge density helps us calculate how much charge is within any section of a sphere, which is critical when using Gauss's law for points inside the distribution.

Electric field inside a sphere

To find the electric field inside a uniformly charged sphere, we use Gauss's law. Here, the radius of our Gaussian surface is smaller than the whole sphere, which gives a fraction of the total charge it contains. The enclosed charge \(Q_{\text{enclosed}}\) within the radius \(r\) of \(0.03 \text{ m}\) is:

• \( Q_{\text{enclosed}} = \rho \times \frac{4}{3} \pi r^3 \)

Using Gauss's law, the equation is simplified for the electric field \(E\):

• \( E \cdot 4 \pi r^2 = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \)
• Solve for \(E\): \( E = \frac{\rho \cdot r}{3\varepsilon_0} \)

This setup elegantly shows how the electric field varies linearly with the radial distance from the center when within the sphere, highlighting the impact of the symmetric charge distribution.

Electric field outside a sphere

For calculating the electric field at a distance outside of a uniformly charged sphere, we apply Gauss's law with another spherical surface, this time encompassing the entire sphere. When entirely outside, the sphere behaves as if all its charge were concentrated at its center. Thus, the electric field \(E\) depends only on the total charge Q and the distance from the center \(r\):

• \( E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \)
• Solving for \(E\), we get: \( E = \frac{Q}{4\pi \varepsilon_0 r^2} \)

Key to remember is, outside the sphere, the field strength decreases with the square of the distance, following an inverse-square law similar to gravitational fields. Thus, at the 6 cm radial distance given in our problem, employing these concepts computes the field strength effortlessly.