Lab 202_ Numerical Verification of Gauss's Law

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New Jersey Institute Of Technology *

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121A

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Mathematics

Date

Apr 3, 2024

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pdf

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9

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Lab 202: Numerical Verification of Gauss's Law Name: Eman Mughal Group ID: 4 Date of experiment: 02/15/2024 Date of report submission: 02/21/2024 Course and Section Number: Phys 121A007- Physics I Lab Instructor’s Name: Diptee Ramakant Tekade Partner’s name: Eman Mughal, Kateryna Sazonova, Deep Shah, Anna Clara
1- Introduction 1.1 Objective The objective of this lab is to utilize MATLAB to numerically analyze Gauss's Law for various surfaces and demonstrate that the integral across the surface is multiplied by ε o , the value of the enclosed charge. 1.2 Theoretical Background The theoretical background of this lab is to figure out how guasses;s works for various surfaces. Electric flux is defined as a dot product of the electric field vector E and the area vector A, with E.A = |E||A|cos(theta). The area vector A of a surface is described as a vector with magnitude (surface area) and direction perpendicular (normal) to it. The standard definition of electric flux is equal to the integral (E. dA), where the equation is the surface integral of the dot product of the electric field (E) and area vectors. According to Gauss's equation, the net flow through any closed surface is integral (E.dA) = Q/e0, where E represents the electric field at any point on the surface and Q represents the net charge inside the surface. 2- Experimental Procedure 2.1 Equipment List There were no equipments used in this experiment. Only MATLAB was used for writing the codes.
Question 1: Question 2
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Question 3 Question 4
Question 5 Question 6.1
Question 6.2 Question 6.3
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Question 7 3- Results 3.1 Results: Data Q1 Q2 Q3 Q4 (x = 0.5, Y = 0.2, Z = 0.7) Q5 Q6.1 a=2 Q6.2 a=10 Q6.3 a=100 Q7 Phi top 169..49 153.8 153.8 350.8 701.7 65.1 3.2 0.032 -165.45 Phi bottom 169..49 153.8 153.8 77.7 155.5 65.1 3.2 0.032 32.4 Phi left 169..49 153.8 101.2 104.8 209.6 65.1 3.2 0.032 34.2 Phi right 169..49 153.8 300.1 155.9 311.8 65.1 3.2 0.032 34.2 Phi front 169..49 300.1 153.8 241.5 483.01 65.1 3.2 0.032 34.2 Phi back 169..49 101.2 153.8 86.05 172.1 65.1 3.2 0.032 34.2
3.2 Results: Calculations This lab didn’t have any calculations. We were only supposed to write codes into MATLAB and put the output values we got from the code into the lab report using the table given by the instructor. 4- Analysis and Discussion The analysis of this lab is that we were able to meet the objective of this lab by testing Gauss's law for various surfaces. We were able to get pretty accurate values and the behavior of the data given also made sense. If the same thing is done in reality, I think there would be a lot of factors like temperature and conductivity that will affect the process. Q1:Place the charge q1 at the center of the cube. Confirm that flux through each surface is the same. The flux is the same through each surface. Q2: . Place the same charge gt on the x axis at position x = 1/2a. What flux is the largest? What flux is the smallest? Are there any values identical? Which ones? Has the total flux & changed? How would you explain that? The flux for the phi front is the largest with 300.170. Phiback has the smallest flux and phi top, down, left, and right has identical flux. The total flux value didn't change because the charge within the surface generate s the same net flux regardless of its location. Q3: Place the same charge on y axis at position y =1/2a. What flux is the largest? What flux is the smallest? Are there any values identical? Which ones? Have the flux & changed? How would you explain that? Phi top and bottom have the same flux. Phi right has the largest flux. Phi left has the smallest flux. The total flux didn’t change because the charge within the surface generate s the same net flux regardless of its location.
Q4:Choose any location inside the cube. How does the flux & change? Explain why. The total flux doesn’t change because any place on the surface the total flux will be the same. It will just be differently distributed. Q5: Double the magnitude of the charge q1. By what factor does the flux change? Explain why. The flux changed by the factor of 2 because we multiplied the magnitude of the charge q 1 by 2. Q6 : Increase the size of the cube to a = 2, then a = 10, then a = 100. Does the flux change? Does Gauss's law appear to hold? Explain. According to Gauss's law, area is inversely proportional to the total flux. Due to the gauss’s law hold, the bigger the size the smaller the flux is going to be. Q7 : Change the size of the cube back to 1 and place a positive charge on the z axis at z = 2a (a charge is placed outside of the cube). How does the total flux change? Explain why. The total flux is zero because the charge is no longer on the surface so it causes no impact on it. 5- Conclusion The conclusion of this lab is that w e were able to accurately evaluate Gauss's Law numerically for various surfaces and confirm that this definition is equal to the enclosed charge value.
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