What is the Electrostatic Force on Particle 3 in a Triangular Configuration?

[jmckennon]

Homework Statement

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).

Homework Equations

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

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

theta= arctan(y/x)

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 haven't 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.

 

[Redbelly98]

Homework Statement

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).

Homework 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)

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 haven't 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 -.

 

[baba89]

I understand your frustration with not being able to get the correct answer. It is important to carefully review your calculations and make sure you are using the correct units throughout. Additionally, it may be helpful to break down the problem into smaller components and solve them individually before combining them to get the final answer.

For this specific problem, I would suggest using the equation F = kQ1Q2/r^2 to calculate the force between particles 1 and 3, and then between particles 2 and 3 separately. This will give you two components of the total force on particle 3. Then, using vector addition, you can combine these two forces to get the net force on particle 3.

I have solved the problem using this method and have obtained the same answers as the book: (a) (0.829N)i and (b) (-0.621N). I have attached a diagram to help visualize the problem and my solution. I hope this helps and good luck with your future calculations!