Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
Does the trend in the F vs r graph correspond to what you should expect from Coulomb’s Law? (b) Of the three relationships (F vs 1/r, F vs 1/r^2, or F vs 1/r^3), which one gives the best linear relationship? Use your linear trendline and R^2 values as evidence. (c) Assuming the F vs 1/r^2 relationship is a linear relationship, use Coulomb’s Law (Equation) to find an expression of the slope of the graph of FEvs 1/r^2. The expression should be in terms of constants (i.e. not r or FE). What should be the units of the slope of the graph?

Does the trend in the F vs r graph correspond to what you should expect from Coulomb’s Law? (b) Of the three relationships (F vs 1/r, F vs 1/r^2, or F vs 1/r^3), which one gives the best linear relationship? Use your linear trendline and R^2 values as evidence. (c) Assuming the F vs 1/r^2 relationship is a linear relationship, use Coulomb’s Law (Equation) to find an expression of the slope of the graph of FEvs 1/r^2. The expression should be in terms of constants (i.e. not r or FE). What should be the units of the slope of the graph?

icon
Related questions
Question

(a)  Does the trend in the F vs r graph correspond to what you should expect from Coulomb’s Law?

(b) Of the three relationships (F vs 1/r, F vs 1/r^2, or F vs 1/r^3), which one gives the best linear relationship? Use your linear trendline and R^2 values as evidence.

(c) Assuming the F vs 1/r^2 relationship is a linear relationship, use Coulomb’s Law (Equation) to find an expression of the slope of the graph of FEvs 1/r^2. The expression should be in terms of constants (i.e. not r or FE). What should be the units of the slope of the graph?

(d) Using the expression you derived for the slope of the graph and the numerical slope you got from the trendline of the FE vs 1/r^2 graph, find the value of ke. Compare it with the theoretical value of ke.??=|?????―??ℎ????| /??ℎ????  × 100You should not just “plug in” a pair of data points and use an equation.

(e) Does Coulomb’s Law seem to be a good model for the electric interaction between two point charges? Use your data and analysis as evidence. Regardless of your response, give sources of uncertainty and error in this experiment.

r(m) F(nc) Force vs radius 0.201 0.000312 0.00035 0.21 0.000280 0.22 0.000261 0.0003 0.231 0.000235 0.00025 0.241 0.000216 0.0002 ..... 0.252 0.000195 0.00015 0.261 0.000186 0.269 0.000173 0.0001 0.281 0.000158 0.00005 0.29 0.000145 0.05 0.1 0.15 0.2 0.25 0.3 0.35 r(m) 1/r (1/m) F(nc) Chart Title 4.98 0.000312 0.00035 4.76 0.000280 y = 0.0001x - 0.0002 R = 0.9958 4.55 0.000261 0.0003 4.33 0.000235 0.00025 4.15 0.000216 0.0002 3.97 0.000195 0.00015 3.83 0.000186 3.72 0.000173 0.0001 3.66 0.000158 0.00005 3.45 0.000145 1 2. 4 1/r (1/m) (ɔu)-
Transcribed Image Text:r(m) F(nc) Force vs radius 0.201 0.000312 0.00035 0.21 0.000280 0.22 0.000261 0.0003 0.231 0.000235 0.00025 0.241 0.000216 0.0002 ..... 0.252 0.000195 0.00015 0.261 0.000186 0.269 0.000173 0.0001 0.281 0.000158 0.00005 0.29 0.000145 0.05 0.1 0.15 0.2 0.25 0.3 0.35 r(m) 1/r (1/m) F(nc) Chart Title 4.98 0.000312 0.00035 4.76 0.000280 y = 0.0001x - 0.0002 R = 0.9958 4.55 0.000261 0.0003 4.33 0.000235 0.00025 4.15 0.000216 0.0002 3.97 0.000195 0.00015 3.83 0.000186 3.72 0.000173 0.0001 3.66 0.000158 0.00005 3.45 0.000145 1 2. 4 1/r (1/m) (ɔu)-
1/r^2 (1/m^ F(nc) 24.8 0.000312 Chart Title 22.7 0.000280 0.00035 20.7 0.000261 y = 1E-05x - 2E-06 R2 = 0.9984 0.0003 18.7 0.000235 0,00025 ... 17.2 0.000216 0.0002 15.7 0.000195 I 0.00015 ...0......... 14.7 0.000186 13.8 0.000173 0.0001 12.7 0.000158 0.00005 11.9 0.000145 10 15 20 25 30 1/r^2 (1/m^2) 1/r^3 (1/m^|F(nc) 123.1 0.000312 Chart Title 108 0.000280 0.00035 94 0.000261 y = 2E-06x + 7E-05 R2 = 0,9949 0.0003 81.1 0.000235 71.4 0.000216 0.00025 62.5 0.000195 0.0002 ............................. . 56.4 0.000186 0.00015 •..... 51.4 0.000173 45.1 0.000158 0.0001 41 0.000145 0.00005 20 40 60 80 100 120 140 (ɔu)
Transcribed Image Text:1/r^2 (1/m^ F(nc) 24.8 0.000312 Chart Title 22.7 0.000280 0.00035 20.7 0.000261 y = 1E-05x - 2E-06 R2 = 0.9984 0.0003 18.7 0.000235 0,00025 ... 17.2 0.000216 0.0002 15.7 0.000195 I 0.00015 ...0......... 14.7 0.000186 13.8 0.000173 0.0001 12.7 0.000158 0.00005 11.9 0.000145 10 15 20 25 30 1/r^2 (1/m^2) 1/r^3 (1/m^|F(nc) 123.1 0.000312 Chart Title 108 0.000280 0.00035 94 0.000261 y = 2E-06x + 7E-05 R2 = 0,9949 0.0003 81.1 0.000235 71.4 0.000216 0.00025 62.5 0.000195 0.0002 ............................. . 56.4 0.000186 0.00015 •..... 51.4 0.000173 45.1 0.000158 0.0001 41 0.000145 0.00005 20 40 60 80 100 120 140 (ɔu)
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 5 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer