Place position (7') values in meters in column A, 2 in column B, and force (F) in T column C. All of these values should be SI base units. 1.P|ace column labels in cell 1. Start placing values in cell 2. 2. In Column B, you are merely multiplying the constant k by the inverse square of column A: = 8.99E09 /A2"2. Copy the formula down. 3. Plot F 0.3. 7'. Make sure you insert the correct graph! Label the graph for presentation. Afterwards, run a trendline by right-clicking the data. Chose power t. Place the equation on the graph. Note the power it t the curve with. 4. Plot F '03. 9%. Is it linear? Label the graph for presentation. Run a LINEST by highlighting a 2 x 5 matrix starting around a15 or so. The theoretical slope should be the product of the two charges. So the slope should have units of 02. Note the value of the slope and its uncertainty. 5. In the summary tab, address these questions: - From the data (you) collected, does the power t indeed illustrate the inverse- square Law? Explain. I We don't know the charge on either sphere. But, we DO KNOW the product. Determine the product of the charges. Does your result seem reasonable? I List all random error in this lab that you can identify. Would these account for any discrepancies for the power t? r(m) k/r Force (F) 0.37 6.5668E+10 0.00025 0.31 9.3548E+10 0.00029 0.223 1. 8078E+1: F vs r 0.00039 0. 153 3.8404E+11 0.0008 0.0 08 0.104 8. 3118E+11 0.00139 0.0 07 0.076 1.5564E+12 0. 00183 0.0 06 0.057 2.767E+12 0. 00241 0.0 05 0.045 4.4395E+12 0. 00298 Force (N) 00 04 0.0339 7.8228E+12 0.00357 y = 0.0001x 1 031 0.026 1.3299E+13 0.00412 0.0 03 0.016 3.51176+13 0.00484 0.0 02 0.0 01 1.2612E-16 0.00131601 005 01 015 0.2 025 1035 3.0196E-17 0.00035285 r (meters) 0. 65965048 0.00100115 17.4434048 9 1.7484E-05 9.0207E-06 F vs k/r2 0.0 06 0.0 05 0.0 01 Force (N 0.0 0: 070 02 0.0 01 SE+ 12 16+ 13 15 E+13 26+ 13 k/r2 Yes this graph is linear Slope 1. 26E-16