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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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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