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

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The discussion revolves around solving a physics problem involving two point charges and a particle in an electric field. The user is attempting to calculate the mass of a particle released from rest, but their answer differs from the book's solution. Key calculations involve determining the net force acting on the particle due to the two charges, with particular attention to the signs of the charges. Despite confirming the calculations and unit conversions, the user and another participant arrive at the same result, suggesting a possible typo in the book. The conversation highlights the complexity of electric force calculations and the importance of careful consideration of charge signs.
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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
 
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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 -
 
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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
 
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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.
 
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