Exercise 1. What is this magnitude if instead of meters in Figure 1 we used nanometers? Place your work in the cell below.

In [1]:THE CODE IS BELOW

x_s, y_s = [1e-9,1e-9] # source coordinates x_t, y_t = [3e-9,3e-9] # target coordinates k_e = 8.99e9 # N m^2 / C^2 q_s = -1.60217662e-19 # C q_t = q_s Fmag = k_e*abs(q_t*q_s)/((x_t-x_s)**2+(y_t-y_s)**2) print('The Magnitude of the Coulomb Force between two electons separated by sqrt(8) meters is = %e N' % Fmag)

SHOULD PRINT THIS: The Magnitude of the Coulomb Force between two electons separated by sqrt(8) meters is = 2.884632e-11 N 

In [3]

# plot the force vector between a pair of charges TEST CELL

import numpy as np # math helper routines

import matplotlib.pyplot as plt # graphics helper routines

x_s, y_s = [1e-9,1e-9] # source coordinates

x_t, y_t = [3e-9,3e-9] # target coordinates

q_t = q_s

k_e = 8.99e9

q_s = -1.60217662e-19 

Fmag = k_e*abs(q_t*q_s)/((x_t-x_s)**2+(y_t-y_s)**2)

plt.plot(x_s, y_s, 'ro', markersize=12)

plt.text(x_s-.2, y_s+.1,'$q_s$',fontsize=16)

plt.plot(x_t, y_t, 'gs', markersize=12)

plt.text(x_t+.1, y_t-.1,'$q_t$',fontsize=16)

plt.arrow(x_s,y_s,x_t-x_s,y_t-y_s,width = 0.06, length_includes_head = True)

plt.text(2.1,1.75,'$\\vec F(s,t)$',fontsize=16)

plt.title('Figure 1',fontsize=18)

plt.grid('on')

plt.xlabel('x (m)', fontsize=12)

plt.ylabel('y (m)', fontsize=12)

plt.axis('equal');

PLEASE USE PYTON