Problem 3 - University Physics with Modern Physics

Question

22.3 You measure an electric field of \(1.25 \times 10^{6} \mathrm{N} / \mathrm{C}\) at a distance of 0.150 \(\mathrm{m}\) from a point charge. (a) What is the electric flux through a sphere at that distance from the charge? (b) What is the magnitude of the charge?

Step-by-Step Solution

Answer

(a) The electric flux is \(4.23 \times 10^{5}\,\mathrm{Nm^2/C}\). (b) The magnitude of the charge is \(3.74 \times 10^{-6}\, \mathrm{C}\).

Step 1: Understand the problem

We have a point charge creating an electric field of \(1.25 \times 10^{6}\,\mathrm{N/C}\) at a distance of 0.150 m. We need to find the electric flux through a sphere of radius 0.150 m and the charge's magnitude.

Step 2: Apply Gauss's Law for the Electric Flux

According to Gauss's Law, the electric flux \(\Phi_E\) through a closed surface is given by \(\Phi_E = \frac{q}{\varepsilon_0}\), where \(q\) is the charge inside the surface, and \(\varepsilon_0\) is the permittivity of free space \((\varepsilon_0 = 8.85 \times 10^{-12}\,\mathrm{C^2/Nm^2})\). For a sphere surrounding a point charge, the electric flux is the same as \(E \cdot 4\pi r^2\).

Step 3: Calculate Electric Flux

We compute \(\Phi_E = E \cdot 4\pi r^2\). Substituting the given values, \(E = 1.25 \times 10^{6}\,\mathrm{N/C}\) and \(r = 0.150\,\mathrm{m}\), we find:\[\Phi_E = 1.25 \times 10^{6} \cdot 4\pi (0.150)^2\]\[\Phi_E = 1.25 \times 10^{6} \cdot 4\pi \cdot 0.0225\]Solving this gives us \(\Phi_E = 4.23 \times 10^{5}\,\mathrm{Nm^2/C}\).

Step 4: Apply Gauss's Law to Find the Charge

Now use Gauss’s Law to solve for the charge: \(q = \Phi_E \cdot \varepsilon_0\). Substitute the calculated \(\Phi_E\) and \(\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{C^2/Nm^2}\):\[q = 4.23 \times 10^{5} \cdot 8.85 \times 10^{-12}\]This yields \(q \approx 3.74 \times 10^{-6}\, \mathrm{C}\).

Key Concepts

Gauss's Law

Gauss's Law is a fundamental principle of electromagnetism connecting electric fields and charges. It states that the total electric flux \(\Phi_E\) through a closed surface is equal to the charge \(q\) enclosed divided by the permittivity of free space, \(\varepsilon_0\). The equation is expressed as:

• \( \Phi_E = \frac{q}{\varepsilon_0} \)

For symmetrical charges like point charges, Gauss's Law greatly simplifies calculations of electric fields. You can envision Gauss's Law as a way to estimate the number of electric field lines passing through a surface.
Let's consider a spherical surface around a point charge. By symmetry, the electric field has the same magnitude \(E\) across the sphere's surface, which leads us to another form of the law:

• \( \Phi_E = E \cdot A \)

Here, \(A\) is the surface area of the sphere, \(4 \pi r^2\).
Combining the expressions, we know the electric flux also equals \( E \cdot 4 \pi r^2 \). In problems like the one described, Gauss's Law helps find unknown quantities, such as charge, lending ease to complex calculations.

Point Charge

A point charge is an idealized model of a charged object. It is considered to be infinitely small, allowing all its charge to be concentrated at one point.
In practice, this helps simplify calculations by avoiding detailed geometric considerations.
Point charges create electric fields that radiate out evenly in all directions, forming spherical symmetry. This simplification is particularly helpful when applying Gauss's Law, as the fields are easy to calculate. The electric field \(E\) generated by a point charge can be calculated using the equation:

• \( E = \frac{k \cdot |q|}{r^2} \)

Where \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \mathrm{Nm^2/C^2}\), \(q\) is the charge of the point charge, and \(r\) is the distance from the charge.
Since the electric field of a point charge diminishes with the square of the distance from the charge (as expressed in the equation above), you can imagine an inverse-square relation to its strength, which is a common occurrence in physical laws involving fields spreading in three dimensions.

Electric Field

The electric field is a vector field surrounding electric charges. It represents the force that a charge would experience at any given point within the field.
The strength and direction are given by the force per unit charge, so the electric field \(E\) is expressed as:

• \( E = \frac{F}{q} \)

Where \(F\) is the force experienced, and \(q\) is the charge present.
Electric fields can be created by point charges or configurations of multiple charges, and they play a critical role in determining the behavior of electric forces.
In the exercise example, the electric field experienced at a distance of 0.150 m from a point charge was given, allowing us to calculate the electric flux through a surrounding sphere and the charge itself using Gauss's Law. The radial nature of electric fields from point charges is fundamental in applying symmetrical assumptions in our calculations.