Capacitors Lab Yip

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Arizona State University *

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Feb 20, 2024

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1 (1 point) Title of the Experiment: Capacitors Lab Student’s name: Puurich Yip Section SLN: 13710 TA’s Name: Sang-Eon Bak, Francesco Setti Week of the experiment: 4 (lab 3)

2 OBJECTIVE ( 3 points ): The purpose of this experiment is to learn and prove how parallel plate capacitors function, and how they react when put into series and parallel circuits. We will also see the dielectric constant being put into series and find out how to find the equivalent capacitance. EXPERIMENTAL DATA ( 6 points ) & DATA ANALYSIS & RESULTS ( 10 points ) Obtain experimental data that will be used for further calculations from the graphs or tables. Be sure to show your calculations and to include related equations and diagrams! PART I. Parallel plate capacitor. a. Air-filled parallel plate capacitor • The labeled screen capture of the air-filled virtual capacitor designed in PhET • SIMULATION WITH V1=1.101V, K=1, d=6.6mm, A=232.5mm^2 V 1 = 1.101 (Volts) [See Module 4 on Canvas for V 1 value] 𝛆 = 𝛋 𝐚?𝐫 ∙ 𝛆 ? (??𝐭? 𝛋 𝐚?𝐫 = ? 𝐚?𝐝 𝛆 ? = 𝟖. 𝟖?? ? ?? −?? 𝐅 ? )

3 C = capacitance; Q = plate charge; U = stored energy d 1 = 6.6 (mm); A 1 = 232.5 (mm 2 ) [See Module 4 on Canvas for d 1 and A 1 values] CALCULATIONS FOR C, Q, U VALUES, WITH PERCENT ERRORS • How do the calculated values of C, Q and E compare with the corresponding ones shown in the simulation – comment on this and refer to the % errors calculated? All three of the calculated values of C, Q, and E came out to be very close to the simulated values, within 0.6% error for the C and E values, and 2% error for the Q value. If the simulation were to give more decimals for the values, there would ’ ve been even less error. I also wasn ’ t able to get the values exactly to the set parameters, however they ’ re within the allowed ranges. Simulated Calculated (show formula and calculations) % error C (unit) 0.31*10^-12F 0.312*10^-12F 0.6% Q (unit) 0.35*10^-12C 0.343*10^-12C 2% U (unit) 1.90*10^-13J 1.89*10^-13J 0.5%

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4 b. Changing d and A • First row: original d and A values from part I-a. Second and third row: change d while keeping original A value. Fourth and fifth row: change A while keeping original d value. [See Module 4 on Canvas for d 2 , d 3 , A 2 , and A 3 values] Answer the question: • What are the effects of changing the plate separation and the plate surface area on the capacitance, the amount of stored charges and energy. The capacitance decreases with an increase in plate separation (d). The energy values and stored charges also decrease with increased separation. With an increase in area, the capacitance, energy, and stored charges also increase. c. With dielectric – changing offset value ( x ) • The labeled screen capture of the dielectric-filled virtual capacitor designed in PhET V 1 = 1.101 (Volts). 𝛆 𝐃 = 𝛋 ∙ 𝛆 ? (with 𝛋 = 4.5); d 1 = 6.5 (mm); A 1 = 232.5 (mm 2 ) [See Module 4 on Canvas for 𝛋 value] When the offset x = 0: C D = 1.42 *10^-12F Q D = 15.61 *10^-13C U D = 8.59 *10^-13J d (mm) A (mm 2 ) C (unit) Q (unit) U (unit) d 1 = 6.6 A 1 = 232.5 0.312*10^-12F 0.343*10^-12C 1.89*10^-13J d 2 =6.7 A 1 = 232.5 0.307*10^-12F 0.338*10^-12C 1.86*10^-13J d 3 =6.9 A 1 = 232.5 0.298*10^-12F 0.328*10^-12C 1.81*10^-13J d 1 =6.6 A 2 = 236.4 0.317*10^-12F 0.349*10^-12C 1.92*10^-13J d 1 =6.6 A 3 = 240.4 0.322*10^-12F 0.355*10^-12C 1.95*10^-13J

5 SIMULATION SHOWING NO CHANGE IN A, V, OR d, WITH A DIELECTRIC CONSTANT 4.5 Answer the questions: • By how much are the values of C D , Q D , and U D different from the corresponding numbers registered for your custom air-filled capacitor for the same parameters of d, A, and V? The values are increased by a multiplier of about 4.5 for C, Q, and U. This is due to the dielectric constant increase by the same multiplier. The k constant scaled the values up by 4.5 as a result of the change. • Do your measurements validate the statement that inserting a dielectric between the plates increases the capacitance by a factor equal to its dielectric constant? The measurements validate the statement that inserting a dielectric between the plates increases the capacitance by a factor equal to its dielectric constant. In my original air-filled capacitor, a value 0.312*10^-12F was found for capacitance. With the same area, voltage, and separation, adding a dielectric with a constant of 4.5, the capacitance came out to be 1.42*10^-12F. This value is approximately 4.5 times our original.

6 d. Capacitance C D vs. Offset  x for a paper-filled capacitor Consider the range of x between 0 -15 mm [See Module 4 on Canvas for ∆ 𝑥 1 , ∆ 𝑥 2 , ∆ 𝑥 3 , ∆ 𝑥 4 , and ∆ 𝑥 5 values]  x (unit) ∆𝑥 0 = 0 ∆𝑥 1 = 3 ∆𝑥 2 = 6 ∆𝑥 3 = 9 ∆𝑥 4 = 12 ∆𝑥 5 = 15 C D (unit) 1.10*10^-12F 0.94*10^- 12F 0.79*10^- 12F 0.63*10^- 12F 0.48*10^- 12F 0.33*10^- 12F • LoggerPro – graph of C vs. Δ x for a paper-filled capacitor, with linear fit. LOGGERPRO GRAPH OF C VS. DELTA X, WITH THE B VALUE SHOWN. y-intercept b as obtained from the linear fit = 1 Δb = 1.871*10^-3pF

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7 CALCULATIONS OF K, DELTAK, AND PERCENT ERROR OF K PART II. Capacitors in parallel – sharing charges • The labeled screen capture of the capacitors in parallel designed in PhET

8 SIMULATION OF CIRCUIT SHOWING V, C1, C2, AND C3. V 0 = 7.0 (Volts), C 1 = 0.12 (F), C 2 = 0.12 (F), C 3 = 0.09 (F). [See Module 4 on Canvas for V 0 , C 1 , C 2 , and C 3 values] 1- Charging C 1 - W 1 closed, W 2 and W 3 open V 1 (measured) = 7V Q 1 = C 1 *V 1 = .12*7C = .84 F Q 21 = 0 (C 2 not charged so far) The magnitude of positive charges deposited on C 1 = .12F The magnitude of negative charges deposited on C 1 = .12F 2- Sharing the charge stored on C 1 with capacitor C 2 – W 2 closed; W 1 and W 3 open V 2 (measured) = 3.5V Q 12 = C 1 *V 2 =.12*3.5= .42C Q 2 = C 2 *V 2 =.12*3.5 = .42C Q 32 = 0 (C 3 not charged so far) - Is the measured voltage V 2 different then V 1 ? Why? Yes, the measured voltage V2 is different because when the second switch is closed, half of the voltage is shared from C1 to C2. - Q i1 = Q 1 + Q 21 = C 1 *V 1 + 0 = C 1 *V 1 Q f2 = Q 12 + Q 2 = C 1 * V 2Theory + C 1 * V 2Theory = (C 1 + C 2 )* V 2Theory Use the fact that Q f2 = Q i1 to solve for V 2Theory = …………………………… V2theory= C1*V1 / C1+ C2 = 0.84/0.24= 3.5V - Does your measured voltage V 2 agree with the calculated theoretical value of V 2Theory ? Yes, the measured and theoretical values are the same. 3- Sharing the charge stored on C 2 with capacitor C 3 – W 3 closed; W 1 and W 2 open V 3 (measured) = 2 V Q 23 = C 2 *V 3 = .12*2=0.28C Q 3 = C 3 *V 3 = .09*2=.21C

9 - Q i2 = Q 2 + Q 32 = C 2 *V 2 + 0 = C 2 *V 2 = .12*3.5=.42C Q f3 = Q 23 + Q 3 = C 2 * V 3Theory + C 3 * V 3Theory = (C 2 +C 3 )* V 3Theory Use the fact that Q f3 = Q i2 to solve for V 3Theory = V3Theory = C2*V2 /(C2+C3)= 0.42/.21=2V - Does your measured voltage V 3 agree with the calculated theoretical value of V 3Theory ? - Yes, the values are the same. - Which capacitor stores less charges and why? - Capacitators 2 and three store less charge than 1 because C1 shares its charge with C2, which then shares with C3. 4- Distribution of charges between three capacitors in parallel – W 2 and W 3 closed; W 1 open. V 4 (measured) = 2.54 V; - Q i3 = Q12 + Q23 + Q3 = C 1 *V 2 + C 2 *V 3 + C 3 *V 3 = (0.12*3.5)+(0.12*2)+(0.9*2)=.84C(C); Q f 4 = C 1 * V 4Theory + C 2 * V 4Theory +C 3 * V 4Theory = (C 1 + C 2 +C 3 )* V 4Theory Use the fact that Q f4 = Q i3 to solve for V 4Theory = V4Theory = Qi3 / (C1 + C2 +C3) = 0.84/0.33= 2.54V - Does your measured voltage V 4 agree with the calculated theoretical value of V 4Theory ? Yes, they agree, the values match. PART III. Combination of capacitors in series and in parallel V 0 = 7 (Volts), C 1 = .12 (F), C 2 = .16 (F), C 3 = .11 (F), C 4 = .20 (F). 1- Charging C 1 - W 1 closed, W 2 open V 1 (measured) = 7V Q 1 = C 1 *V 1 = 0.84C Q EQ1 = 0 (C 2 , C 3 , C 4 not charged so far) 2- Sharing the charge stored on C 1 with capacitor C 2 – W 2 closed; W 1 open V 12 (measured across C 1 ) = 4.97V V 2 (measured across C 2 ) = 1.53V V 3 (measured across C 3 ) = 2.22V V 4 (measured across C 4 ) = 1.22V

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10 SIMULTION SHOWING V OF C4= 1.22V Q 12 = C 1 * V 12 = .12*4.97=0.60C V EQ = V 12 Q EQ2 = Q 1 - Q 12 = C 1 *V 1 - C 1 *V 12 = C 1 *(V 1 - V 12 ) = .12*(7-4.97)=0.24C C EQ2 = Q EQ2 / V EQ = 0.24/4.97=0.048F C EQ,Theo = (C2*C3*C4)/(C2*C3+C3*C4+C2*C4)= (.16*.11*.20)/(.16*.11+.11*.20+.16*.20)=0.049 % error of C EQ = (.048-.049)/(.048) *100= 2% - According to what you have learnt about capacitors connected in series what charge should be stored on each of the three capacitors C 2 , C 3 , C 4 ? ( Theoretical prediction) Q 2 = C 2 * V 2 = .16*1.53 = 0.24C Q 3 = C 3 * V 3 = .11*2.22 = 0.24C Q 4 = C 4 * V 4 = .20*1.22 = 0.24C - Do the experimentally determined charges Q 2 , Q 3 , Q 4 agree with the theoretical prediction? (Hint: find the charge on an equivalent capacitor that experiences the same potential drop as the series capacitors. How is the charge on this equivalent capacitor related to the charge on each of the series capacitors?) Q 2 = .24C Q 3 = .24C Theoretical Experimental

11 Q 4 = .24C DISCUSSION & CONCLUSION ( 10 points ): This experiment was conducted to determine and prove how capacitive plates, dielectric materials, and capacitive discharge operate. Part 1 was to dive into the capacitive plates. We played around with the distance and areas of these plates, and also changed the dielectric constant. We found that an increase in distance between the plates caused a decrease in capacitance, energy values, and energy stored. We also found that an increase in area causes an increase in capacitance, energy values, and energy stored. With the dielectric constant, we found that it will increase the capacitance by whatever factor the dielectric constant is. In our case, the dielectric constant was 4.5. This increased our capacitance by a factor of 4.5. We also used paper as our dielectric material. Theoretically paper has a dielectric constant of 3.5. My measured value was 3.2, which gave an error value of 3.2%. In part 2, we dove into circuits, and showed how capacitors behave in parallel circuits. When in parallel, we found that once charged capacitors share their charges with open circuits. In our experiment, we found that C1 shared with C2, however only C2 would share with C3. C1 cannot share charges with C3. The theoretical values and measured values were strongly consistent with one another. Part 3 of this experiment was about series circuits, which we found that capacitors share a voltage discharge equivalent to the capacitive value. In our case, we found that C2, C3, and C4 were all sharing the 4.97V, which was shared from C1. The theoretical values were proven true with measurements.

Capacitors Lab Yip

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