# Title Lab Number and Title: Lab 202: Numerical Verification of

Ultimate Gauss’ Law review
Ultimate Gauss’ Law review

Lab 202 – Gauss Law – Lab Report

## University

New Jersey Institute of Technology

## Course

### Preview text

Title Lab Number and Title: Lab 202: Numerical Verification of Gauss’s Law Name: Abhinit Sundar Date of Experiment: 10/4/ Course & Section Number: Physics II Lab – PHYS 121A – 003 Partners’ Names: Kriss Sitapara, Nifesimi Akintola, Jhaylor Cudia Group ID: Date of Report Submission: 10/9/ Instructor’s Name: Matias Daniel de Almeida

1. Introduction

1 Objective

a. In this section, using MATLAB software, you will evaluate Gauss’s Law numerically for different surfaces and verify that the integral over the surface multiplied by €o is the value of the enclosed charge.

1 Theoretical Background

In the preceding experiment, electric flux was examined after a basic understanding of electrostatic attraction and repulsion. In this experiment, Gauss’s Law was calculated through MATLAB software by taking the integral over the surface area enclosed within the charge and multiplying that by the electric field. Gauss’s Law is in essence the electric field value multiplied by the surface area and a factor of cosine of thetha. Since we are dealing with the dot product, the value for Gauss’s Law will always employ with it cosine of thetha. There are varying levels of area vectors that could theoretically correlate to Gauss’s Law. In other words, the Gaussian

distribution can be any enclosed surface comprising of a variable area vector. A sphere will have a different Gaussian distribution compared to a rectangular prism or a cylinder. Nonetheless, the formula for finding the Gaussian electric flux is the same: the electric flux of that surface multiplied by the surface area of the respective object and a factor of cosine of the angle evaluated. This principle of Gauss’s Law was evaluated by writing MATLAB code to find the electric flux of a cube for all the six faces. Etop denotes the electric field of the top face of the cube while Ebottom, Eleft, Eright, Efront, and Eback denote the electric fields of the bottom, left, right, front, and back faces respectively. Phitop, Phibottom, Phileft, Phiright, Phifront, and Phiback collectively represent the individual electric flux components of the top, bottom, left, right, front, and back portions of the cube, which is the Gaussian distribution in this scenario. The electric flux for each of these individual faces is the respective electric field value, which could be either positive or negative, multiplied by the surface area of the particular component. The total electric flux or Phitotal is best described as the sum of the individual electric fluxes. In this scenario, Phitotal would be the sum of Phitop, Phibottom, Phileft, Phiright, Phifront, and Phiback.

Net Flux Through Any Closed Surface:

E = electric field at any point on the surface Q = net charge inside the surface

Accuracy in computing the resultant total electric flux of the cube which depends on the individual electric flux components

Gauss’s Law was utilized to calculate the electric flux of an enclosed surface for various values of x, y, and z. Throughout this experiment, the same formula of using an integral of the electric field vector multiplied by the derivative of the surface area enclosed was applied. A small area, known as dA, was chosen to calculate the flux d(phi) through the area using the dot product of the electric field vector, E, and the area vector, and integrating through the different portions of the cube. The following underlying formula, , was used throughout this experiment. The only difference was that the x, y, and z values differed in the problems. The individual electric flux was computed for the top, bottom, left, right, front, and back surfaces of the cube. The sum of all these individual electric flux values yields the total electric flux for the Gaussian distribution, which in this case is a cube. Altering the x, y, and z values varied the individual electric flux for the 6 different surfaces of the cube: top, bottom, left, right, front, and back and thereby impacted the total electric flux.

There were not any additional procedures carried out in this experiment than the ones outlined in the laboratory manual for Lab 202: Numerical Verification of Gauss’s Law. 3. Results 3 Experimental Data MATLAB code and output for computing Gauss’s Law

Phitop (electric flux of top surface of cube)

Phibottom (electric flux of bottom surface of cube)

Phiright (electric flux of right surface of cube)

Phifront (electric flux of front surface of cube)

AllPhi (total of all individual Phi, or electric flux – total electric flux of cube)

3 Calculation

1. The electric flux values were equivalent in the first set of calculations where x, y, or z were either a or -a. Since the value of electric flux is equivalent to k multiplied by q divided by r squared, the magnitude for the electric flux for the top, bottom, left, right, front, and back portions were all the same. Also, r is equivalent to the sum of x squared, y squared, and z squared raised to the three-halves power and multiplied by r to the one-half power. Essentially, the formula for electric flux of multiplying the electric constant k by q divided by r squared is constant. Since the value of a was constant across these problems and only the x, y, or z positions were substituted with this value, the electric flux for all 6 components was 577 (Nm^2) / C, thereby confirming that the flux through each surface is the same. The total electric flux, hence, was simply 6 multiplied by the individual electric flux, which was 577 (N * m^2) / C, which yields 3465 (N m^2) / C.

2. In the second set of calculations, x was equivalent to (1/2)a. The top and bottom faces of the cube both carried identical flux values and were the largest in magnitude with a value of 1543 (N*m^2) / C. The left and right faces of the cube both carried identical flux values and were the next largest in magnitude with a value of 789 (N * m^2) / C. The front and back faces of the cube both carried identical flux values and were the smallest in magnitude with a value of 577 (N * m^2) / C.

The top and bottom faces carried the largest flux values since the sum of the square of the x component (1/2 * a), y component, and z component (-a or a) is the largest compared to the other calculations. Also, in both the x and z positions, a value of a was substituted into them, giving greater magnitude to the r component and yielding a larger electric flux value. The value of k multiplied by q is constant in all the scenarios since k is a constant and the same charge was unanimously applied across all the problems in the Analysis section of this laboratory report. The

left and right faces carried lower individual electric flux values since a value of a is only applied once, in the y position. The other positions for the electric field are of the x and z components, which yield lower individual electric flux values. In the front and back electric flux components, the value is the lowest since a is applied in the x component and y and z are used. Since the position of a has changed from the y component to the x component, a lower value is yielded for the electric flux using the Gauss’s Law equation. Ultimately, the value of the electric flux will increase with greater magnitude of a. Since the first case had the greatest magnitude for a, the electric flux was the largest. If the magnitude of a is lower, accordingly, the electric flux will then depend on the position of a. Specifically, it will depend on whether the x, y, or z component is substituted with a value for a.

The total electric flux changed from the first scenario from a value of 3465 (Nm^2) / C to 5821 (Nm^2) / C.

1. In the third set of calculations, y was equivalent to (1/2)a. The top and bottom faces of the cube both carried identical flux values and were the largest in magnitude with a value of 3595 (Nm^2 / C). The front and back faces of the cube both carried identical flux values and were the next largest in magnitude with a value of 789 (Nm^2 / C). The left and right faces of the cube both carried identical flux values and were the smallest in magnitude with a value of 735 (N*m^2 / C). The top and bottom faces carried the largest flux values since the sum of the square of the x component, y component – (1/2)a – and z component (-a or a) is the largest compared to the other calculations. Also, in both the y and z positions, a value of a was substituted into them, giving greater magnitude to the r component and yielding a larger electric flux value. The value of k multiplied by q is constant in all the scenarios since k is a constant and the same

Lab 202 – Gauss Law – Lab Report

You are watching: Title Lab Number and Title: Lab 202: Numerical Verification of. Info created by THVinhTuy selection and synthesis along with other related topics.

Rate this post