Question

Calculate the magnitude of the force between two point charges of +1.6×10^-19 C and -1.6×10^-19 C, for a range of distances, from r = 0.090 nm to r = 0.18 nm in 0.010 nm steps.

Plot the results on a graph of magnitude of force F versus distance r. You must label the graphs with axis labels including units, and a title. Also choose an appropriate scale of the graph to show the data.

Hint: use a spreadsheet, Logger Pro, or write some computer code to do this for you. You must try to do this type of problem on the computer, as writing this out by hand/doing repetitive calculations is just too time-consuming.

Answer 1

From Couloumb's law of electrostatics

the electrostatic force between the two chatges (say q1,q2), separated by a distace r is directly propotional to the product of the charges and inversly proportional to square of the distance between the charges.

    where k is proportionality constant called coulomb's constant

where ε₀ = permittivity of space

now the problem is given charges q1,q2 are +1.6×10^-19 C and -1.6×10^-19 C , separated by distance r = 0.090 nm to r = 0.18 nm,

F= 9*10^9 Nm^2/C^2*1.6×10^-19 C * (-1.6×10^-19 C)/(0.090*10^-9m)*(0.090*10^-9m)

= 28.444*10^-9 N

For the next position of the charges

F= 9*10^9 Nm^2/C^2*1.6×10^-19 C * (-1.6×10^-19 C)/(0.1*10^-9m)*(0.1*10^-9m)

= 23.04*10^-9 N----------------------A

F= 9*10^9 Nm^2/C^2*1.6×10^-19 C * (-1.6×10^-19 C)/(0.11*10^-9m)*(0.11*10^-9m)

=19.04*10^-9 N----------------------B

for r= 0.120 nm F= 16.00*10^-9 N--------C

for r=0.130 nm F= 13.633*10^-9 N--------D

for r = 0.140 nm F= 11.755*10^-9 N-------E

for r= -.150 nm F= 10.24*10^-9 N---------F

for r= 0.16 nm F= 9.0*10^-9 N--------------G

for r= 0.17 nm F= 7.972*10^-9N-------------H

for r= 0.18 nm F= 7.1111*10^-9 N------------I