Problem 9 - Fundamentals of Physics

Question

Fig. 23-31 shows a Gaussian surface in the shape of a cube with edge length \(1.40 \mathrm{~m}\). What are (a) the net flux \(\Phi\) through the surface and (b) the net charge \(q_{\mathrm{cnc}}\) enclosed by the surface if \(\vec{E}=(3.00 y \hat{j}) \mathrm{N} / \mathrm{C},\) with \(y\) in meters? What are (c) \(\Phi\) and (d) \(q_{\text {cnc }}\) if \(\vec{E}=[-4.00 \hat{\mathrm{i}}+(6.00+3.00 y) \mathrm{j}] \mathrm{N} / \mathrm{C} ?\)

Step-by-Step Solution

Answer

(a) \(8.232 \text{ Nm}^2/\text{C}\), (b) \(7.287 \times 10^{-11} \text{ C}\); (c) \(8.232 \text{ Nm}^2/\text{C}\), (d) \(7.287 \times 10^{-11} \text{ C}\).

Step 1: Calculate Flux for Field (a)

Given the electric field \( \vec{E} = (3.00y \hat{j}) \) N/C and the side length of the cube \( L = 1.40 \) m. Since the field is in the \( \hat{j} \) direction, it affects the top and bottom faces of the cube.For the top face at \( y = 1.40 \) m:\[\Phi_{\text{top}} = E \cdot A = (3.00 \times 1.40) \times (1.40)^2 = 3.00 \times 1.40 \times 1.96 = 8.232 \text{ Nm}^2/\text{C}\]For the bottom face at \( y = 0 \) m:\[\Phi_{\text{bottom}} = E \cdot A = (3.00 \times 0) \times (1.40)^2 = 0\]The net flux through the surface is:\[\Phi = \Phi_{\text{top}} - \Phi_{\text{bottom}} = 8.232 - 0 = 8.232 \text{ Nm}^2/\text{C}\]

Step 2: Calculate Enclosed Charge for Field (a)

According to Gauss's Law, the net flux \( \Phi \) is related to the enclosed charge \( q_{\mathrm{cnc}} \) by:\[\Phi = \frac{q_{\mathrm{cnc}}}{\varepsilon_0}\]Plug in the values:\[8.232 = \frac{q_{\mathrm{cnc}}}{8.854 \times 10^{-12}}\]Solve for \( q_{\mathrm{cnc}} \):\[q_{\mathrm{cnc}} = 8.232 \times 8.854 \times 10^{-12} = 7.287 \times 10^{-11} \text{ C}\]

Step 3: Calculate Flux for Field (c)

Given \( \vec{E} = [-4.00 \hat{i} + (6.00 + 3.00y) \hat{j}] \) N/C.The \(\hat{i}\) component does not contribute to flux since it is parallel to the cube faces it intersects.For the top face at \( y = 1.40 \) m:\[\Phi_{\text{top}} = (6.00 + 3.00 \times 1.40) \times 1.96 = 10.20 \times 1.96 = 19.992 \text{ Nm}^2/\text{C}\]For the bottom face at \( y = 0 \) m:\[\Phi_{\text{bottom}} = 6.00 \times 1.96 = 11.76 \text{ Nm}^2/\text{C}\]The net flux through the surface is:\[\Phi = \Phi_{\text{top}} - \Phi_{\text{bottom}} = 19.992 - 11.76 = 8.232 \text{ Nm}^2/\text{C}\]

Step 4: Calculate Enclosed Charge for Field (c)

Using Gauss's Law again:\[\Phi = \frac{q_{\text{cnc}}}{\varepsilon_0}\]Plug in the values computed:\[8.232 = \frac{q_{\text{cnc}}}{8.854 \times 10^{-12}}\]Thus, solving for \( q_{\text{cnc}} \):\[q_{\text{cnc}} = 8.232 \times 8.854 \times 10^{-12} = 7.287 \times 10^{-11} \text{ C}\]

Key Concepts

Electric Flux

Electric flux is a measure of how an electric field interacts with a given surface. Imagine the electric field lines passing through a surface. Electric flux quantifies the number of these lines that go through the surface.

It is given by the formula:

• \( \Phi = \vec{E} \cdot \vec{A} \) where \( \vec{E} \) is the electric field vector and \( \vec{A} \) is the area vector of the surface.

The direction of the area vector is perpendicular to the surface, and its magnitude equals the surface area.

For a closed surface, we consider the net flux which is the total flux through the entire surface. This helps in determining if there is any charge enclosed by the surface.

Electric Field

The electric field is a vector field around a charged object, describing the force a positive test charge would experience. In simpler terms, it is a field that represents the influence a charge has on other charges around it.

The electric field is defined as:

• \( \vec{E} = \frac{\vec{F}}{q} \)

where \( \vec{F} \) is the force experienced by a charge \( q \). It gives us an idea of how a charge moves in the presence of other charges. The units of electric field are \([ \text{N/C} ]\).

In the exercise, the electric field varies with position, represented as \( \vec{E} = (3.00y \hat{j}) \text{ N/C} \). This shows how the electric field strength changes with the distance \( y \) along the \( y \)-axis.

Enclosed Charge

In context with Gauss's Law, the enclosed charge refersto the total charge contained within a closed surface. Gauss'sLaw is a powerful method for calculating the net chargewithin an enclosed surface through the concept of electric flux.

Gauss's Law states that the electric flux \( \Phi \) through a closed surface is directly proportional to the enclosed charge \( q_{\text{enc}} \):

• \( \Phi = \frac{q_{\text{enc}}}{\varepsilon_0} \)

where \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.854 \times 10^{-12} \) \( \text{F/m} \).

It allows us to relate the electric field's behavior over the surface to thequantity of charge within it.

Gaussian Surface

The Gaussian surface is an imaginary surface that we use in applying Gauss's Law. It is strategically chosen to simplify the calculation of the electric field or enclosed charge. The surface can be any shape but is chosen to be symmetric to match the charge distribution as much as possible.

In the cube-shaped Gaussian surface from the exercise, each face of the cube is perpendicular to the electric field lines, which makes calculating electric flux straightforward. With the electric field given, the flux through each surface can be easily determined based on how they align with the field direction.

Using a Gaussian surface is often more theoretical than practical, but it is extremely useful for solving problems involving electromagnetism, saving calculation effort by utilizing symmetry.