In the section above, we used the smallest Ag to determine the elementary charge 6. This would only work if we happened to get two droplets with exactly one 'e' difference in charge not yery likely in reality! Instead, a more realistic option would be to use the smallest common factor of all Ag values. Explain how this allows a better measurement of e rather than relying on the smallest Aq. We will use the following table to measure our value for 'e': Charge (C, Ascending Order) Aq (C) 5.101 x 10-18 5.420*10^-18 3.19*10^-19 Your last answer was interpreted as follows: 5.420 . 10-18 Your last answer was interpreted as follows: 3.19 - 10-19 5.738*10^-18 3.18*10^-19 Your last answer was interpreted as follows: 5.738 . 10-18 Your last answer was interpreted as follows: 3.18 - 10-19 7.970*10^-18 2.23*10^-18 Your last answer was interpreted as follows: 7.970 . 10-18 Your last answer was interpreted as follows: 2.23 . 10-18 8.289*10^-18 3.19*10^-19 Your last answer was interpreted as follows: 8.289 . 10-18 Your last answer was interpreted as follows: 3.19 . 10-19 9.245*10^-18 9.56*10^-19 Your last answer was interpreted as follows: 9.245 - 10-18 Your last answer was interpreted as follows: 9.56 - 10-19 4.47*10^-18 1.371 x 10-17 Your last answer was interpreted as follows: 4.47 . 10-18 Step 1: Fill in the charges in ascending order in the first column; because we will be looking at the differences in charge, it makes the most sense to arrange them with the smallest differences between neighbours. Remember how to enter scientific notation: 1 x 10% would be entered as '1e3'. Step 2: Calculate the difference in charge, Aq, between each pair of measurements (filling in the second column of the table). (Enter values as magnitudes only, don't include sign). Step 3: Use the smallest (non-zero) value for Aq to record your value for 'e' below; 3.18*10^-19 Your last answer was interpreted as follows: 3.18 . 10-19