Question

The electric flux through a spherical Gaussian surface of radius \(R\) centered on a charge \(Q\) is \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right) .\) What is the electric flux through a cubic Gaussian surface of side \(R\) centered on the same charge \(Q ?\) a) less than \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) b) more than \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) c) equal to \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\) d) cannot be determined from the information given

Short answer

Answer: b) more than 1200 N/(Cm²)

Step-by-step solution

Recall Gauss's Law

Gauss's Law states that the electric flux through any closed surface \(S\) is given by the total charge enclosed by the surface divided by the permittivity of free space, which is a constant, \(\varepsilon_0\). Mathematically, it's represented as: $$\Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{encl}}{\varepsilon_0}$$ where \(\Phi_E\) is the electric flux, \(\vec{E}\) is the electric field, and \(d\vec{A}\) is the differential area vector.

Calculate the total charge enclosed for the spherical gaussian surface

For the given spherical Gaussian surface, we have the electric flux as \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\). We will now calculate the total enclosed charge using Gauss's Law: $$Q_{encl} = \Phi_E \cdot \varepsilon_0$$

Calculate the electric field at the surface of the cube

Now, we will calculate the electric field at the surface of the cubic Gaussian surface, which is centered on the same charge \(Q\). The electric field is the same at every point on the surface of the cube. So, we can use the enclosed charge we found in step 2 and calculate the electric field magnitude as follows: $$E = \frac{Q_{encl}}{4\pi R^2 \varepsilon_0}$$

Determine the electric flux through the cubic Gaussian surface

To find the electric flux through the cubic Gaussian surface, we will use Gauss's Law again. Since the electric field is the same at every point on the surface of the cube, we can calculate the electric flux as follows: $$\Phi_E = E \cdot A = \frac{Q_{encl}}{4\pi R^2 \varepsilon_0} \cdot A$$ Now, we need to find the total surface area of the cube. The surface area of one side of the cube is \(R^2\), and there are 6 sides. So, the total surface area is \(A = 6 R^2\). Let's substitute the value of \(A\) in the above equation: $$\Phi_E = \frac{Q_{encl}}{4\pi R^2 \varepsilon_0} \cdot 6 R^2$$ We can simplify the equation as: $$\Phi_E = \frac{3Q_{encl}}{2\pi R^2 \varepsilon_0}$$

Compare the electric flux through the cubic and spherical Gaussian surfaces

Now, we have the expression for the electric flux through the cubic Gaussian surface, and we know the electric flux through the spherical Gaussian surface, which is \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\). We need to compare them to choose the appropriate answer. First, let's notice that the denominator of the cubic surface's flux expression is \(2\pi R^2 \varepsilon_0\), which is exactly half of spherical surface's flux denominator (\(4\pi R^2 \varepsilon_0\)). Since the numerator (\(3Q_{encl}\)) remains the same, the electric flux through the cubic Gaussian surface is 1.5 times greater than the flux through the spherical Gaussian surface. So, the answer is: b) more than \(1200 \mathrm{~N} /\left(\mathrm{C} \mathrm{m}^{2}\right)\)

Key concepts

Gauss's Law

When exploring the concept of electric flux, it is essential to understand the foundational principle known as Gauss's Law. Essentially, Gauss's Law connects the electric flux, which is a measure of the quantity of electric field lines passing through a given area, to the charge enclosed by that area. Mathematically, it is expressed as
\[\Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{encl}}{\varepsilon_0}\]
Here, \(\Phi_E\) is the electric flux through a closed surface \(S\), \(\vec{E}\) is the electric field, \(d\vec{A}\) is the differential area perpendicular to the field, \(Q_{encl}\) is the enclosed charge, and \(\varepsilon_0\) represents the permittivity of free space, a fundamental constant. The beauty of Gauss's Law lies in its simplicity and broad applicability; regardless of the shape of the closed surface—be it spherical, cubic, or any other geometry—the law holds true. A key to understanding problems involving electric flux is recognizing that the total electric flux only depends on the amount of charge within the Gaussian surface, and not on the shape of the surface or the charge distribution.

Gaussian Surface

Let's delve deeper into the concept of a Gaussian surface. A Gaussian surface is an imaginary closed surface used in conjunction with Gauss's Law to simplify the process of calculating electric fields or electric flux. The beauty of choosing an appropriate Gaussian surface lies in its symmetry and its ability to make the calculations feasible.

In the context of the provided exercise, we encounter two Gaussian surfaces: a sphere and a cube, both centered around the same charge \(Q\). When it comes to a spherical surface, the symmetry ensures that the electric field is uniform across the entire surface, which greatly simplifies the calculation of electric flux. On the other hand, a cubic surface may not offer the same level of symmetry, but as Gauss's Law states, the total electric flux will still solely depend on the charge enclosed.

It's important for students to visualize these surfaces and understand why and how they are used. While the actual electric field may vary at different points on a non-spherical surface, the overall flux through the Gaussian surface remains constant as it is proportional to the enclosed charge, a crucial takeaway for solving many electromagnetism problems.

Electric Field

The electric field is a vector quantity that represents the force experienced by a unit positive charge placed at a point in space. It emanates from positive charges and terminates on negative charges, and it is described both by its magnitude and direction. For a charge \(Q\) in free space, the magnitude of the electric field it produces can be calculated using Coulomb's Law:
\[E = \frac{kQ}{r^2}\]
where \(k\) is Coulomb's constant and \(r\) is the distance from the charge.

In our exercise, to determine the electric flux through the cubic Gaussian surface, the calculation involved the electric field \(E\) at the surface of the cube, using the formula derived from Gauss's Law. It’s pivotal for students to comprehend that although the electric field might not be uniformly distributed across a non-spherical Gaussian surface, Gauss's Law allows us to consider the net flux, which simplifies the seemingly complex situation into an approachable calculation.

Moreover, by understanding that the electric field is a measure of how strongly space is 'influenced' by electric charge, students can better conceptualize why the electric flux—essentially the 'flow' of electric field through a surface—would be influenced by both the intensity of the field and the size of the surface through which it passes.