Calculating Net Force on Charged Particles in an Equilateral Triangle

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To calculate the net force on each charged particle in an equilateral triangle formation, first set up a coordinate system with one particle at the origin. Use Coulomb's law (F = k*q1*q2/r^2) to find the forces exerted by the other two particles on the particle of interest. Calculate the x and y components of these forces, ensuring one charge is aligned along the x-axis for simplicity. Add the components separately to determine the net force vector. A review of vector math may be beneficial for clarity in this calculation.
ranger
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I don't know how to approacg this problem.

Lets say that we have 3 positively charged particles, each having an equal charge of 11.0 uC (11*10^-6 C). These three particles are located at the (three) corners of an equilateral triangle with sides 0.15m. How can I calculate the magnitude and direction of the net force on each particle?

There is no need to do the math for me, just explain step-by-step how to approach it.

If anyone could help me out with this, I would appreciate it>>
What is the total charge of all electrons in 1.0kg of water.


--thank you
 
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The total force on one of the particles is the sum of the forces coming from the other two particles. Set up a coordinate system with one of the particles at the center. Then just add up the i and j components from each of the other particles, calculated from coulomb's law, and you're done.
 
ranger said:
What is the total charge of all electrons in 1.0kg of water.
What is the charge on one electron? How many electrons are in each water molecule? How many water molecules are in 1kg of water? Once you answer these questions, just multiply them together.
 
The total force on one of the particles is the sum of the forces coming from the other two particles

How would I find that force? Is it by using the equation:

Code:
F = k*q1*q2/r^2

LEts say that want to find the force that is applied to q3. would finding it look something like this?
Code:
#calculate force from q1
F= k*q1*q3/r^2

#calculate force from q2
F = k*q2*q3/r^2

Then I add the two answers?

Set up a coordinate system with one of the particles at the center. Then just add up the i and j components from each of the other particles, calculated from coulomb's law, and you're done.

Could you please be a little more explanatory on that one.
 
ranger said:
How would I find that force? Is it by using the equation:

Code:
F = k*q1*q2/r^2
Yes, use that equation.
Could you please be a little more explanatory on that one.
Ok, set up an x-y coordinate system. Put the charge your looking at in the origin. Each charge will exert a force with components along these two directions (Although it would be most convenient to make one of the components of one of the forces 0 by lining one of the charges along the x-axis so that it pushes in this direction only and the y-component is 0) Find the x- and y-components of the forces from each of the charges and add up the components seperately. Like I said before, it would be a good idea to have one of the charges sitting on the x-axis. The force from this charge is just the force you calculated from coulomb's law directed along the x-axis(whatch out for the sign: it should point away from the other charge since they repel). The other charge will be at an angle. The component of the force along the x-axis will be F\cos{\theta} where F is the force from Coloumb's law and \theta is the angle between the line joining the charge to the origin and the x-axis. The y-component will be F\sin{\theta}.
 
Thanks for your help. Its seems my main problem is simple vector math. I did that in pervious physics course 1yr ago. I will have to do a review on that. Anyone recommend any goog online reading on the topic.
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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