### Electric Flux Through a Gaussian CubeAn electric field given by \(\vec{E} = 3.0 \hat{i} - 9.6(y^2 + 4.3) \hat{j}\) pierces the 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.)#### Problem Statement:What is the electric flux through the:- (a) Top face,- (b) Bottom face,- (c) Left face,- (d) Back face?(e) What is the net electric flux through the cube?#### Figure Description:The figure shows a Gaussian surface in the shape of a cube. The cube is aligned with the coordinate axes, with one corner at the origin of the coordinate system (x, y, z axes). The cube extends outward from the origin, with its faces parallel to the coordinate planes.- The top face of the cube is parallel to the xy-plane at the highest z-value of the cube.- The bottom face of the cube is also parallel to the xy-plane at the lowest z-value of the cube.- The left face of the cube is parallel to the yz-plane.- The back face of the cube is parallel to the xz-plane. Each face of the cube has an edge length of 0.180 meters.##### Tips for Calculating Electric Flux:Electric flux (\(\Phi\)) through a surface is calculated using the formula:\[\Phi = \int \vec{E} \cdot \vec{dA}\]Where:- \(\vec{E}\) is the electric field vector.- \(\vec{dA}\) is the differential area vector, which is perpendicular to the surface.Consider the contributions of each component of the electric field for the respective faces of the cube to determine the flux through that face. The net electric flux through the cube can be computed by summing up the flux through each face.Remember, according to Gauss's law for a closed surface, the net electric flux through a closed surface is proportional to the net charge enclosed within the surface.

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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? 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? 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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