What Is the Mass of a Particle in an Electric Field Problem?

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Homework Help Overview

The problem involves calculating the mass of a particle in an electric field created by two fixed point charges. The scenario includes a positive and a negative charge, with the particle released from a specific position and experiencing an initial acceleration.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations of forces acting on the particle, questioning the net force and the inclusion of charge signs. There are attempts to clarify the forces involved and the correct application of formulas.

Discussion Status

Participants are actively engaging with the problem, raising questions about the calculations and the interpretation of forces. Some guidance has been offered regarding the consideration of charge signs, but there is no explicit consensus on the correct answer yet.

Contextual Notes

There is mention of a discrepancy between the calculated mass and the answer provided in the textbook, leading to discussions about potential typos or errors in the book. Participants are also addressing the conversion of units for acceleration.

rob24
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hey guys, i have been trying to work this problem out but for some reason, my answer is different from the books answer on the back.
the problem is as follow:
Two points charges of 30 nC and -40nC are held fixed on an X axis, at the origin and at X=72 cm, respectively. A particle with a charge 42 micro C is released from rest at X=28 cm. if the initial acceleration of the particle has a magnitude of 100 Km/s^2, what is the particle's mass?

the way i was solving it:
let q1=30nC and q2= -40nC q3=42 Mic C X=72 cm
F13 = (8.99*10^9*(30*10^-9)*(42*10^-6))/(28*10^-2)^2= 0.144
F23 = (8.99*10^9*(40*10^-9)*(42*10^-6))/(44*10^-2)^2= 0.078
F12 = (8.99*10^9*(30*10^-9)*(40*10^-9))/(72*10^-2)^2=2.08*10^-5

so Fnet= -F13+F23+F12 and then F=ma so m=F/a.

Am i solving it incorrectly. Please Help. Thanks
 
Last edited:
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Fnet on what? What's the net force acting on the third charge (42 microC)?
 
neutrino said:
Fnet on what? What's the net force acting on the third charge (42 microC)?
I did that too. it's F13 and F23. so Fnet without F12 but it's still wrong.
 
Are you sure you've taken into consideration the nature of the charges?
 
yes. please tell me if you have different idea.
 
rob24 said:
yes.
I don't think so. You haven't included the minus sign (-) of the negative charge whlie calculating F23.
 
you are correct but the answer is still incorrect. i had added the - sign later after calculation. so i had - and -
 
Last edited:
What's you answer? And, btw, did you also convert Kms-2 to ms-2?
 
I kept getting 2.2*10^-6 but in the book he has it as 2.2*10^-5.
of course i had 100km/s^2 as 100000m/s^2
 
  • #10
EDITED
Oops, sorry...too late into the night for calculations. I'm getting the same answer as you. May be it's a typo in the book.
 
Last edited:

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