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Electrostatic force question

by jmckennon
Tags: electrostatic, force
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jmckennon
#1
Sep11-08, 02:13 PM
P: 42
1. The problem statement, all variables and given/known data
Three charged particles form a triangle: particle 1 with charge Q1=80nC is at xy coordinates (0,3.00mm) particle 2 with charge q2 is at (0,-3.00mm) and particle 3 is at(4.00mm,0). In unit vector notation, what is the electrostatic force on particle 3 due to the other two particles if Q2 is equal to (a) 80.0 nC and (b) -80nC?

The book lists the answers as: (a).(0.829N)i and (b) (-.621N).




2. Relevant equations

F= (|Q1||Q2|*K)/r^2

k=permittivity constant 8.99x10^9
r= distance between the charges

theta= arctan(y/x)
3. The attempt at a solution

I have tried countless different ways to do it and can't seem to get the book's answer. I have played around with the numbers for over 3 hours and havent found anything that works. Any help would be GREATLY appreciated as I have no idea how close or far away I am.

The last way I tried it was by drawing out the triangle. I found theta to be arctan(3/4)=36.869degrees

F=((.64x10^24)*8.99x10^9)/(6x10^-3)^2=159.822 nC/m^2

Fx=159.822cos(36.869)=-127.851
Fy=159.822sin(36.869)=95.89132
Fy= 159.822

sqrt((159.822+95.89132)^2+(-127.851)^2))=285.8973

I have no idea what the units are there after all that on the latest try, nor what that answer means.
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Redbelly98
#2
Sep11-08, 03:15 PM
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Quote Quote by jmckennon View Post
1. The problem statement, all variables and given/known data
Three charged particles form a triangle: particle 1 with charge Q1=80nC is at xy coordinates (0,3.00mm) particle 2 with charge q2 is at (0,-3.00mm) and particle 3 is at(4.00mm,0).
What is the charge of particle #3?

In unit vector notation, what is the electrostatic force on particle 3 due to the other two particles if Q2 is equal to (a) 80.0 nC and (b) -80nC?

The book lists the answers as: (a).(0.829N)i and (b) (-.621N).




2. Relevant equations

F= (|Q1||Q2|*K)/r^2
What does this have to do with Q3, the particle you are asked to calculate the force for?

k=permittivity constant 8.99x10^9
r= distance between the charges

theta= arctan(y/x)
3. The attempt at a solution

I have tried countless different ways to do it and can't seem to get the book's answer. I have played around with the numbers for over 3 hours and havent found anything that works. Any help would be GREATLY appreciated as I have no idea how close or far away I am.

The last way I tried it was by drawing out the triangle. I found theta to be arctan(3/4)=36.869degrees
Yes.

F=((.64x10^24)*8.99x10^9)/(6x10^-3)^2=159.822 nC/m^2
q1 q2 = (.64x10^24) -- Use Q3 instead of Q2. Is the exponent really +24?
k = 8.99x10^9 -- Yes
r = (6x10^-3) -- No. You need the distance between q1 and q3, not between q1 and q2.

Fx=159.822cos(36.869)=-127.851
Fy=159.822sin(36.869)=95.89132
This will be correct once you calculate F correctly. Just keep track of whether these components are + or -.


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