D2L-Exp4

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Apr 3, 2024

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Capacitator Lab: Basics Professor Dhiraj Maheswari February 12 th , 2024

Purpose The purpose of this lab is to investigate the relationship between capacitance of a parallel capacitor and the area of the plates, the distance between plates, and the charge and voltage. Lastly, this lab will exemplify the permittivity of free space and compare the experimental and the accepted value. Introduction Capacitance is defined as the ability of a system to store charge. Inside of a capacitor, two pieces of conducting material are separated from each other in order to store this charge. The pieces will separately maintain the equal but opposite magnitude of charge. The capacitance of a capacitor is the number that tells you how well that capacitor stores charge. In general, capacitance equals the charge stored on a capacitor divided by the voltage across that capacitor. Procedure 1. Open the simulation using the link to PhET Interactive Simulations at the University of Colorado Boulder: PhET Interactive Simulations 2. Explore the options available in the simulator to change area of the plates and distance between the plates of a parallel plate capacitor (see screenshot below). 3. In this lab activity you will be using the Capacitance tab, but feel free to review the Light Bulb options. Data and Evaluation Part 1: Capacitance and Area C= ↋ ₀ A/d: In this part the distance, d, between the plates is kept constant d= 8.0 x10 -3 m and the area of the plates is changing. Record the values for area (in m 2 ) and the capacitance C (in F). Take at least eight values of A and C, and then fill the table below: d= 8.0 x10 -3 m A (m 2 ) C (F)

.0001 .11 x 10 .12 .00014 .15 x 10 .12 .00018 .20 x 10 .12 .00021 .23 x 10 .12 .00023 .25 x 10 .12 .00035 .39 x 10 .12 .0004 .44 x 10 .12 1- Use the Vernier Graphical Analysis tool to plot the relationship between A and C. 2- Draw the best straight-line equation and determine its slope 3- From the slope, determine the value of the permittivity of free space ↋ ᵒ . ↋ ᵒ = 1.112 x 10 -9 x 8.0 x10 -3 m=8.896 x 10 -12 4- Determine the percentage error using the real value ↋ ᵒ =8.85X10 -12 C 2 /N.m 2 (8.85X10 -12 -8.896 x 10 -12 )/ 8.85X10 -12 =.0052x100=.52%

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Part 2: Capacitance and distance C= ↋ ᵒ A/d: In this part the area A, of the plate is kept constant A= 370 x10 -6 m 2 and the distance d between the plates is changed. You are to record the values for distance (in m) and the capacitance C (in F). Take at least eight values of d and C, and then fill the table below by calculating (1/d): A= 370 x10 -6 m 2 d (m) 1/d ( m -1 ) C (F) .005 m 200 6.6 x 10 .13 .0062 161 5.3 x 10 .13 .0066 151 5.0 x 10 .13 .0078 128 4.2 x 10 .13 .0092 108 3.6 x 10 .13 .0098 102 3.3 x 10 .13

.01 100 3.3 x 10 .13 1- Plot the relationship between (1/d) and C. 2- Draw the best straight-line equation and determine its slope.

3- From the slope, determine the value of the permittivity of free space ↋ ᵒ ↋ ᵒ =(.003316x10 -12 )/( 370 x10 -6 )=8.96x10 -12 C 2 /N.m 2 4- Determine the percentage error using the real value ↋ ᵒ= 8.85 x10 -12 C 2 /N.m 2 (8.96x10 -12 - 8.85 x10 -12 ) / (8.85 x10 -12 )=.0124 x 100= 1.24% Part 3: Capacitance and Charge & Voltage: CV=Q 1- Connect the circuit and by changing the area A and the distance between the plates d, set the value for the capacitance C = 0.35 pF. 2- Change the value of the power supply V, then record the values for the charge Q (in C) and the potential difference V ( in V) Capacitance from PhET C = 0.35 pF V(Volts) Q(C) -1.500 -.525 -1.00 -.35

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-.600 -.21 0 .00 .350 .1225 .550 .1925 .750 .2625 1.5 .525 1- Plot the relationship between Q and V, using the Vernier Graphical Analysis tool, and determine the slope.

2- From the slope determine the value of C ( in pF) m= .35 The capacitance is not only given in the problem, but by comparing the equation of the line with Q=CV, the capacitance also equals .35 pF. 3- Calculate the percentage error in your result. (.35-.35)/.35=0x100=0% Results and Conclusion Given the results, it is exemplified that If all other factors are equal, the greater plate area gives greater capacitance and less plate area gives less capacitance. This describes a direct or proportional relationship. However, the distance between plates in a capacitor inversely affects its capacitance. As the distance increases, capacitance decreases, exemplifying an inverse relationship. In part one as well as part two, the value of the slope and the plate separation is used to calculate the permittivity of the free space. In part three, the relationship between capacitance of parallel plates and charge and voltage.