Question

You measure an electric field of 1.25 \(\times\) 10\(^6\) N/C at a distance of 0.150 m from a point charge. There is no other source of electric field in the region other than this point charge. (a) What is the electric flux through the surface of a sphere that has this charge at its center and that has radius 0.150 m? (b) What is the magnitude of this charge?

Short answer

(a) The electric flux is approximately \(3.534 \times 10^5 \text{ Nm}^2/\text{C}\). (b) The charge is approximately \(3.13 \times 10^{-6} \text{ C}\).

Step-by-step solution

Understanding Electric Flux and Gauss's Law

The electric flux through a closed surface is given by Gauss's Law which states that the total electric flux \( \Phi_E \) through a closed surface is equal to the charge \( Q \) enclosed divided by the electric constant \( \varepsilon_0 \): \( \Phi_E = \frac{Q}{\varepsilon_0} \). Because the surface is a sphere centered around the charge, the electric field is uniform over the surface, allowing \( \Phi_E = E \times A \) where \( E \) is the electric field and \( A \) is the surface area.

Calculating the Surface Area of the Sphere

The surface area \( A \) of a sphere is calculated using the formula \( A = 4\pi r^2 \). Given that \( r = 0.150 \) m, we calculate: \[ A = 4\pi (0.150)^2 \approx 0.2827 \text{ m}^2. \]

Calculating the Electric Flux

Given the electric field \( E = 1.25 \times 10^6 \text{ N/C} \), and using the surface area calculated in Step 2, the electric flux \( \Phi_E \) is: \[ \Phi_E = E \times A = (1.25 \times 10^6 \text{ N/C}) \times 0.2827 \text{ m}^2 \approx 3.534 \times 10^5 \text{ Nm}^2/\text{C}. \]

Calculating the Charge using Gauss's Law

Using Gauss's Law, with \( \Phi_E = \frac{Q}{\varepsilon_0} \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2 \), we rearrange to solve for the charge \( Q \): \[ Q = \Phi_E \times \varepsilon_0 = 3.534 \times 10^5 \text{ Nm}^2/\text{C} \times 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2 \approx 3.13 \times 10^{-6} \text{ C}. \]

Key concepts

Electric Flux

Electric flux is a measure of the electric field passing through a given surface. It's an important concept in electromagnetism and is defined by Gauss's Law. To visualize this, think of electric flux as the number of electric field lines going through a surface. If there are more lines passing, the electric flux is greater. The formula for electric flux (\( \Phi_E \)) is \( \Phi_E = E \times A \), where \( E \) is the electric field and \( A \) is the surface area the field is permeating. When dealing with closed surfaces such as spheres, Gauss's Law simplifies the analysis by relating the electric flux to the charge enclosed: \( \Phi_E = \frac{Q}{\varepsilon_0} \). This equation indicates that the total electric flux through a closed surface depends directly on the enclosed charge. In spherical symmetry, the electric field is uniform across the surface, making these calculations straightforward.

Electric Field

The electric field is a vector field that surrounds electric charges and exerts forces on other charges within the field. It tells us both the direction and magnitude of the force experienced by a unit positive charge placed in the field. The formula for the electric field (\( E \)) is \( E = \frac{F}{q} \), where \( F \) is the force experienced and \( q \) is the charge. Typically, in the vicinity of a point charge, the electric field decreases with the square of the distance from the charge. In the scenario we are exploring, the electric field is given as \( 1.25 \times 10^6 \text{ N/C} \). This value indicates how strong the force would be per unit charge at a specific point from our charge.

Point Charge

A point charge is an idealized model of a charged particle in which the entire charge is considered to be concentrated at a single point in space. This simplification allows us to easily calculate electric fields and potentials in the surrounding space by ignoring any physical dimension of the charge. In reality, all charges occupy some volume, but for theoretical purposes, point charges are extremely useful. One remarkable property of point charges is their symmetry; they create a radially outward electric field that simplifies many calculations under theoretical conditions. For example, the electric field (\( E \)) at a distance \( r \) from a point charge \( Q \) is given by \( E = \frac{kQ}{r^2} \), where \( k \) is Coulomb's constant.

Surface Area of a Sphere

The surface area of a sphere is crucial when calculating electric flux, especially in symmetric charge distributions. A sphere is a simple geometric shape where every point on the surface is equidistant from the center. The formula to find the surface area (\( A \)) of a sphere is \( A = 4\pi r^2 \), where \( r \) is the radius of the sphere. This formula helps in finding the surface through which the electric field lines pass. In our case, the radius is \( 0.150 \text{ m} \), leading us to calculate a surface area of approximately \( 0.2827 \text{ m}^2 \). This calculation is significant in determining the electric flux, as a larger surface area would accommodate more field lines, thus increasing the flux.