### Electric Fields and Flux Through a Gaussian Surface**Problem Statement:**An electric field given by \( \vec{E} = 3.0i - 9.6(y^2 + 4.3)j \) pierces a Gaussian cube of edge length 0.180 m and is positioned as shown in the figure. (The magnitude \( E \) is in newtons per coulomb, and the position \( x \) is in meters.) **Questions:**- (a) What is the electric flux through the top face?- (b) What is the electric flux through the bottom face?- (c) What is the electric flux through the left face?- (d) What is the electric flux through the back face?- (e) What is the net electric flux through the cube?**Diagram Explanation:**The diagram illustrates a cube referred to as a Gaussian surface. The cube is depicted with an edge length of \( 0.180 \) meters. The directions of the electric field components \( \vec{E} \) are indicated by unit vectors \( i \) and \( j \), corresponding to the x and y axes.**Answering the Questions:**1. To compute the electric flux through each face, the integral form of Gauss's Law is applied:\[ \Phi_E = \int \vec{E} \cdot d\vec{A} \]where \( \Phi_E \) is the electric flux, \( \vec{E} \) is the electric field, and \( d\vec{A} \) is the differential area vector perpendicular to the surface.2. By evaluating the given \( \vec{E} \) for each specific face's coordinate and integrating over the face area, observations can be made regarding the electric field components normal to each face and thereby determine the corresponding flux.For the provided example:- **Top Face (y = 0.180 m):** - Given \( \vec{E} = 3.0i - 9.6(y^2 + 4.3)j \) - At y = 0.180 m, substitute \( y = 0.180 \) in the electric field expression.- **Bottom Face (y = 0 m):** - Follow the same process but use \( y = 0 \).- **Left Face (x =

Giventhe length of Gaussian cube is 0.18 m then area of each face is given by the electric field…

An electric field given by E-3.0-9.6(y² + 4.3) pierces the Gaussian cube of edge length 0.180 m and positioned as shown in the figure. (The magnitude E is in newtons per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube? (a) Number -8e-6 Units N•m^2/C -Gaussian surface

An electric field given by E-3.0-9.6(y² + 4.3) pierces the Gaussian cube of edge length 0.180 m and positioned as shown in the figure. (The magnitude E is in newtons per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube? (a) Number -8e-6 Units N•m^2/C -Gaussian surface

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

### Electric Fields and Flux Through a Gaussian Surface

**Problem Statement:**

An electric field given by \( \vec{E} = 3.0i - 9.6(y^2 + 4.3)j \) pierces a Gaussian cube of edge length 0.180 m and is positioned as shown in the figure. (The magnitude \( E \) is in newtons per coulomb, and the position \( x \) is in meters.)

**Questions:**

- (a) What is the electric flux through the top face?
- (b) What is the electric flux through the bottom face?
- (c) What is the electric flux through the left face?
- (d) What is the electric flux through the back face?
- (e) What is the net electric flux through the cube?

**Diagram Explanation:**

The diagram illustrates a cube referred to as a Gaussian surface. The cube is depicted with an edge length of \( 0.180 \) meters. The directions of the electric field components \( \vec{E} \) are indicated by unit vectors \( i \) and \( j \), corresponding to the x and y axes.

**Answering the Questions:**

1. To compute the electric flux through each face, the integral form of Gauss's Law is applied:
\[ \Phi_E = \int \vec{E} \cdot d\vec{A} \]
where \( \Phi_E \) is the electric flux, \( \vec{E} \) is the electric field, and \( d\vec{A} \) is the differential area vector perpendicular to the surface.

2. By evaluating the given \( \vec{E} \) for each specific face's coordinate and integrating over the face area, observations can be made regarding the electric field components normal to each face and thereby determine the corresponding flux.

For the provided example:

- **Top Face (y = 0.180 m):**
- Given \( \vec{E} = 3.0i - 9.6(y^2 + 4.3)j \)
- At y = 0.180 m, substitute \( y = 0.180 \) in the electric field expression.

- **Bottom Face (y = 0 m):**
- Follow the same process but use \( y = 0 \).

- **Left Face (x =

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