Particle Physics: Net Force on Charge q1 Calculation

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Homework Help Overview

The discussion revolves around calculating the net force on a point charge (q1 = 2.0 nC) due to two other point charges (q2 = -3.5 nC and q3 = 5.0 nC) positioned along the positive x-axis. Participants are exploring the application of Coulomb's law and the principle of superposition in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Coulomb's law and the superposition principle to determine the forces acting on charge q1. There are attempts to calculate the forces exerted by q2 and q3, with some confusion regarding the correct values and signs in the calculations. Questions about the meaning of "n" in the context of charge units arise, leading to clarifications about nanocoulombs.

Discussion Status

The discussion has seen various attempts to calculate the forces, with some participants successfully recalculating their results after clarifying unit conversions. There is a mix of correct and incorrect calculations being shared, and while some participants express uncertainty, others provide guidance and corrections.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an ongoing exploration of the definitions and implications of the terms used in the calculations.

Amelia
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Homework Statement


Three point charges are located on the positive x-axis of a coordinate system. Charge q1 = 2.0 nC is 2.0 cm from the origin, charge q2 = -3.5 nC is 4.0 cm from the origin and charge q3 = 5.0 nC located at the origin.

What is the net force ((a)magnitude and (b) direction) on charge q1 = 2.0 nC exerted by the other two charges?.[/B]

Homework Equations


F= k (q1q2/r^2)

The Attempt at a Solution


I know that the direction (b) is 0 degrees as all the points are all located on the x-axis, I just can't figure out part (a). I keep mixing up my diagram so that when the equation is applied, my answer comes out incorrectly. Any help would be great! [/B]
 
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Superposition of two forces. Do them one at a time.
 
Bystander said:
Superposition of two forces. Do them one at a time.
Okay, so would it be...
F of 2 on 1 = 9*10^-6((-3.5*10^-6)*(-2*10^-6)/(0.04)^2 = 39.375 N (+ve therefore forces repel)
F of 3 on 1 = 9*10^-6((2*10^-6)*(-5*10^-6)/(0.02)^2 = -225N (-ve therefore forces attract)
F total = 39.375 - (-225) = 264.375 N
 
"n" means what? Like charges attract or repel?
 
Bystander said:
"n" means what? Like charges attract or repel?
'N' in the final result was for Newtons
 
Lower case "n," as in "nC."
 
Bystander said:
Lower case "n," as in "nC."
nC = nanocoulomb
 
"Nano" means what?
 
Bystander said:
"Nano" means what?

ohhh... 1*10^-9 not 1*10^-6 thank you I didn't even see that !
so the new solutions come out at
F of 2 on 1 = -0.000225 N
F of 3 on 1 =0.000039 N
So the final answer would be -0.000225 N - 0.000039 N = -2.46*10^-4 I just tried to enter that in and it still came out incorrect... I'm just not sure what I'm doing wrong
 
Last edited:
  • #10
Amelia said:
F of 2 on 1
4-2=?
 
  • #11
Bystander said:
4-2=?
4-2=2 but I'm not sure where that fits in..
 
  • #12
Amelia said:
F of 2 on 1 = 9*10^-6((-3.5*10^-6)*(-2*10^-6)/(0.04)^2 = 39.375
 
  • #13
Okay so it should be
F of 2 on 1 = 9*10^9((-3.5*10^-9)*(-2*10^-9)/(0.02)^2 = 0.000158 N
F of 3 on 1 = 9*10^9((2*10^-9)*(-5*10^-9)/(0.02)^2 = -0.00025N
Therefore, Ft= 0.000158-(-0.00025) = 3.8*10^-4 N

Which I just put in and it's correct! Thank you so much for your help! :) :)
 
  • #14
You are welcome. Things to remember: one step at a time; and, pay attention to details.
 
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