Question

Consider a uniform nonconducting sphere with a charge \(\rho=3.57 \cdot 10^{-6} \mathrm{C} / \mathrm{m}^{3}\) and a radius \(R=1.72 \mathrm{~m}\). What is the magnitude of the electric field \(0.530 \mathrm{~m}\) from the center of the sphere?

Short answer

The magnitude of the electric field at a distance of 0.530 m from the center of the uniformly charged nonconducting sphere is 212.99 N/C.

Step-by-step solution

Define a Gaussian surface inside the sphere

We will define our Gaussian surface to be a concentric sphere with the same center as the charged sphere, but with a radius of equal to the distance from the center to the point of interest, \(r = 0.530\mathrm{~m}\).

Apply Gauss's law to the Gaussian surface

Gauss's law states that \(\oint \vec{E}\cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}\). Here, \(Q_{enc}\) is the charge enclosed by the Gaussian surface, and \(\vec{E}\) is the electric field at a point on the Gaussian surface. We'll determine the charge enclosed by the Gaussian surface and the area of the Gaussian surface.

Compute the charge enclosed

Since \(\rho\) is constant, we can find the charge enclosed \(Q_{enc}\) by taking the integral of the charge density over the Gaussian surface: \begin{align*} Q_{enc} = \rho \cdot V = \rho \cdot \frac{4}{3}\pi r^3 \end{align*} Now plug in the values for \(\rho=3.57 \cdot 10^{-6} \mathrm{C/m^3}\) and \(r=0.530 \mathrm{~m}\): \begin{align*} Q_{enc} = (3.57 \cdot 10^{-6}\mathrm{C/m^3})\cdot \frac{4}{3}\pi (0.530\mathrm{~m})^3 = 2.554 \times 10^{-6} \mathrm{C} \end{align*}

Find the electric field

Now return to Gauss's law and solve for the electric field: \begin{align*} \oint \vec{E}\cdot d\vec{A} &= \frac{Q_{enc}}{\varepsilon_0}\\ E \oint dA &= \frac{Q_{enc}}{\varepsilon_0} \end{align*} Since the electric field is parallel to the area vector at every point on the Gaussian surface, their dot product becomes a simple multiplication. Also, the electric field is uniform over the Gaussian surface, so it can be pulled out of the integral: \begin{align*} E \cdot 4\pi r^2 &= \frac{Q_{enc}}{\varepsilon_0}\\ \end{align*} Now, solve for \(E\) and plug in the values for \(Q_{enc}\), \(r\), and \(\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}\): \begin{align*} E &= \frac{Q_{enc}}{4\pi \varepsilon_0 r^2}\\ E &= \frac{2.554 \times 10^{-6}\mathrm{C}}{4\pi \cdot 8.85 \times 10^{-12}\mathrm{F/m} \cdot (0.530 \mathrm{m})^2} = 212.99 \mathrm{N/C} \end{align*}

State the final answer

The magnitude of the electric field at a distance of \(0.530 \mathrm{~m}\) from the center of the uniformly charged nonconducting sphere is \(212.99 \mathrm{N/C}\).

Key concepts

Electric Field

The electric field is a vector field around a charged object where an electric force would be exerted on other charged objects. Intuitively, you can think of it as a way to visualize the strength and direction of the electrical force that a charge would experience in space. The unit of electric field is Newtons per Coulomb (N/C).

• The direction of the electric field is the direction of the force that a positive test charge would feel placed in the field.
• The magnitude of the electric field can be calculated using Coulomb's Law, but can also be elegantly determined using Gauss's Law, which can be particularly useful for symmetrical charge distributions.

By calculating the electric field at a specific point, we can understand the impact of a charge distribution in its vicinity. This is essential in predicting how other charged particles will behave when brought into the field.

Uniform Charge Distribution

In the context of electrostatics, a uniform charge distribution signifies that the charge density, typically denoted by \( \rho \), is constant throughout the given volume. This means every segment of the object carries the same amount of charge per unit volume. For example, the sphere mentioned in the problem is uniformly charged with a density of \( \rho = 3.57 \times 10^{-6} \mathrm{C/m^3} \).

• Uniform charge distributions are often assumed in physics problems because they simplify calculations significantly.
• They help apply Gauss's Law effortlessly, since symmetry arguments can simplify the integral calculations.

This concept is foundational for textbooks because many real-world applications assume uniform charge distributions for simplification, even though perfect uniformity may not exist in all practical cases.

Gaussian Surface

A Gaussian surface is an imaginary closed surface used in Gauss's Law to calculate electric fields. This surface is ideally chosen to exploit symmetry, making the integral calculations straightforward. In our problem, the Gaussian surface is a sphere of radius \( r = 0.530 \mathrm{m} \), centered at the same point as the charged sphere.

• The choice of the Gaussian surface often depends on the symmetry of the charge distribution, such as spherical, cylindrical, or planar.
• In the exercise, the Gaussian surface is used to determine the charge enclosed and hence the electric field inside the sphere.

Choosing the right Gaussian surface can significantly simplify the application of Gauss's Law, making it easier to find electric fields for various symmetrical objects.

Electrostatics

Electrostatics deals with the study of charges at rest and the forces, fields, and potentials associated with them. Central to electrostatics is Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed by that surface. This law is particularly powerful for calculating fields around symmetrical charge distributions.

• In electrostatics, charges are assumed to be stationary, and as a result, the electric field they create is constant over time.
• Internet problems often involve using the concept of field lines, field strength, and potential to understand how charges interact with their environment.

Understanding electrostatics is crucial for tackling a wide range of problems in fields like electronics, materials science, and even cosmology, as it sets the foundational understanding of how charges influence their surroundings.