Important Lab 1_ Electric Field (2)

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Rutgers University *

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Physics

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Dec 6, 2023

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LAB 1 Title: Electric field Names: Mario Almonte, Aiman Afzal, Matias De Benedetto, Shehroze Khan Date: 10 Feb. 2022 Abstract: Electric charges apply Force to one another. Magnitude of the force is proportional to charges and inversely proportional to the distance between them, according to Coulombs Law. An alternative way of thinking is that when a charge is present it produces Electric field around it. When another charged object enters the field, it interacts with the field and a force is applied. The objective of this lab is to investigate the Electric field produced by charges and verify the results of Coulomb’s Law. Part A: Open https://phet.colorado.edu/en/simulation/charges-and-fields Click on Grid and Values. Using the ruler tool verify that the distance between each major grid is 0.5 meters and distance between each minor grid is 0.1 meters.
Click and drag one nanocoulomb (1 nC) positive charge near the center of the field. The location of this charge is your origin with coordinates x=0 and y=0 Drag the yellow sensor tool and measure Electric field intensity on the x-axis with the increments of 0.5meters. Record them on the table below Take one screenshot image for Procedure A. Include this image in your lab report. Procedure A DATA Procedure A Theory Calculations and Graphs E = kq/r 2 E is the intensity of the Electric field in N/C or V/m, q is the charge in Coulomb, r is the distance between the field point and source point in meters and k is the universal Coulomb constant. With SI units theoretically k=8.99 10 9 N m 2 /C 2 . Plot E vs 1/r2 and apply a linear trend line. Show the equation and R 2 value. Write your equation and compare with the theoretical equation (calculate % error). E=kq/ r (m) E (N/C) 1/r^2 E(N/C) 0.5 36.7 4.00 36.7 1.00 8.90 1.00 8.90 1.50 4.00 0.4444 4.00 2.00 2.24 0.25 2.24 2.50 1.43 0.16 1.43 3.00 1.00 0.1111 1.00 r (m) E (N/C) 0.50 36.7 1.00 8.9 1.50 4 2.00 2.24 2.50 1.43 3.00 1 3.50 0.73 4.00 0.56
3.50 0.73 0.0816 0.73 4.00 0.56 0.0625 0.56 Part A Calculation: E=Kq/r^2 k=8.99x10^9 Y=mx So, m= slope of Ers 1/r^2 =kq Kex=slope/q Slope=9.1786x So, Kex=9.1786/10^-9 = 9.1786x10^9 So, %error= |kTh –Kexp|/KTh x100 % = |8.99-9.1786| x10^9 / 8.99x10^9 x100
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