PHYS 102 | Coulomb's Law 6 - Use That Unit Vector!
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- Опубликовано: 14 мар 2017
- You can avoid all the trigonometry if you use the definition of the unit vector. I show you how here, but you probably won't need this method for most problems.
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This video blew my mind lol. I've been struggling with this topic for a couple of days because of that unit vector. Every time I search electric forces and fields, the videos I come across only discuss them in terms of magnitude, so I do not get the whole picture. It's all about components! This cleared up my main confusion in under 9 minutes. Brilliant.
You are great professor, thank you!
Thank you it all makes so much more sense now
Thank you! This was helpful
good explaination
i dont understand? -40j would mean that the positive charges experiences a force towards the other positive charge. They repel eachother right?
This form seems to be the most useful if one is going to incorporate the resultant force calculation into a software program.
'''
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)
Professor Hefner, I would like to ask what that i and j hat even mean. Also in the last problem, you left the answer as -2.28 i hat - 0.57 j hat. Why is that so?
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
ruclips.net/video/4lFVUoDEcYE/видео.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 !