VI-1# V1 (Volts) V2 (Volts) p= 1 – V2/V11 19 0 12 32 1 0.968753 18 0.3 0.983334 27 0.1 0.996305 9 0 1 Sample calculation of p:p = 1- V2/V1 = 1 – 0/19 = 1The mean and standard deviation of p values can be calculated using excelp=¿ 0.989676 σp=¿0.013557Calculation of standard deviation of the mean can be done using the following equation:σx=σx√NThus, σp=σp√N=0.013557√5=0.0060628747p± σx=0.9897 ± 0.006VI-2# Vblue (Volts) Vwhite (Volts) Vfinal (Volts) Vblue + Vwhite (Volts)1 0.4 0.1 0.1 0.42 -4.2 5.5 -1.0 0.33 -8.2 8.0 -2.1 -2.34 -5.6 5.8 -1.2 -1Sample calculation of the sum of the reading of two charge producersVblue + Vwhite = 0.4VBased on the conservation of charge, only negative and positive charges can be paired. The Vfinal is the similar to the sum of Vblue and Vwhite since the net charge of an isolated system remained constant.VI-3# Vinsert (Volts) Vground (Volts) Vwithdr (Volts) Vfinal (Volts)1 6.6 0.1 -6.4 0.32 0.9 0.1 -0.8 0.23 1.5 0.1 -1.2 0.4Diagrams of charge distribution for different situations are drawn on the paper at the back of the lab reportVI-4# of transfersnvoltage (volts)5 0.45 -4.25 -8.25 -5.610 0.110 5.510 810 5.815 0.115 -115 -2.115 -1.220 0.420 0.320 -2.320 -125 0.425 0.325 -2.325 -10 5 10 15 20 25 30-10-8-6-4-20246810f(x) = 0.04 x − 0.98V vs. number of charge transfers # of charge transfers Pail voltage (V)# of transfersnvoltage (volts)5 05 05 05 010 010 010 010 015 015 015 015 020 020 020 020 025 025 025 025 00 5 10 15 20 25 3000.10.20.30.40.50.60.70.80.91f(x) = 0V vs. n# of charge transfersShield voltage