PHYS 102 | Coulomb's Law 6 - Use That Unit Vector! 

''' Coulomb's Law Force Calculations in Two Dimensions (x, y) ''' import numpy as np # Given: K_SUB_E = 8.9875517923e9 # Coulomb's constant in N * m^2 / C^2 q1 = q2 = q3 = q4 = 10.0e-6 # Four charges in C r21vec = np.array([0., -0.15]) # Vector r from particle 2 to particle 1 [m, m] r31vec = np.array([-0.6, -0.15]) # Vector r from particle 3 to particle 1 [m, m] r41vec = np.array([-0.6, 0.]) # Vector r from particle 4 to particle 1 [m, m] # 2-dimensional vector magnitude calculation: vecmag = lambda vec: np.sqrt(vec[0]*vec[0] + vec[1]*vec[1]) def elec_f(arg_qpa, arg_qpb, arg_rvec): ''' Electrical force vector calculation: arg_qpa: magnitude of charge of particle A arg_qpb: magnitude of charge of particle B arg_rvec: vector from particle A to particle B ''' r = vecmag(arg_rvec) return K_SUB_E * arg_qpa * arg_qpb * arg_rvec / (r * r * r) # Derivations: F21 = elec_f(q2, q1, r21vec) print('F21:', F21) F31 = elec_f(q3, q1, r31vec) print('F31:', F31) F41 = elec_f(q4, q1, r41vec) print('F41:', F41) F1 = F21 + F31 + F41 print('Resulting F1:', F1)

Those are the unit vectors for the x axis (i-hat) and y (j-hat). They just mean that magnitude in x or y. I left the answer there becuase the problem is basically done at that point. A question might ask for the components or it might ask you to get the total magnitude and angle. Here's a bit on unit vectors ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4lFVUoDEcYE.htmlsi=VYn-y_2Vd6rIvRPM&t=324

So, if I comprehended you correctly, i hat and j hat imply magnitude in x and y direction respectively. If they are negative like (-i hat) and (-j hat), then that mean the magnitude is in the negative x and y co-ordinator plane. In a nutshell, -2.28 i hat and -0.57 j hat mean (-2.28, -0.57), right, Mr. Hefner?

Yes, that is correct.@@andrewjustin256

@@Prof-Hafner Thank you Professor!! You have been so helpful. I love your videos !