How Do You Calculate the X Component of Force in Coulomb's Law?

  • Thread starter Thread starter nateastle
  • Start date Start date
  • Tags Tags
    Law
AI Thread Summary
To calculate the x component of the force on charge q3, the force between charges q2 and q3 must be considered, along with the negative x component from the interaction between charges q1 and q3. The force on the y-axis has been correctly calculated as -3.45 x 10^-5 N. The x component of the total force is approximately 5.4 x 10^-5 N in the positive direction. Proper application of Coulomb's law is essential for accurate results.
nateastle
Messages
15
Reaction score
0
I believe this is the right place for this question. I am strugling with getting this homework problem setup. The question is:

A charge q1 = -3 nc is placed at the origin of an xy-cordinate system, and a charge q2=2.00nC is placed on the positive y Axis at y=4cm. If a third charge q3 =4nC is now placed at point x=3 cm, y=4cm, find the x and y componets of the total force exerted on this charge by the other two.

I have setup my grid and figured out the force on the y Axis is -3.45 X 10^-5 (or at least I hope that is correct) but I am having problems figuring out how to setup the problem to get the force on the x axis. If I didn't calculate the force right for the y-axis please let me know. I have been using coulombs law of f=k(q1q2/r^2)
 
Physics news on Phys.org
nateastle said:
I believe this is the right place for this question. I am strugling with getting this homework problem setup. The question is:

A charge q1 = -3 nc is placed at the origin of an xy-cordinate system, and a charge q2=2.00nC is placed on the positive y Axis at y=4cm. If a third charge q3 =4nC is now placed at point x=3 cm, y=4cm, find the x and y componets of the total force exerted on this charge by the other two.

I have setup my grid and figured out the force on the y Axis is -3.45 X 10^-5 (or at least I hope that is correct) but I am having problems figuring out how to setup the problem to get the force on the x axis. If I didn't calculate the force right for the y-axis please let me know. I have been using coulombs law of f=k(q1q2/r^2)
Your answer for the force on the y-axis is correct, in Newtons. The force in the +x direction is the force between charge 2 and 3 less the x component of the force between 1 and 3 (which is in the - x direction). I get about 5.4 x 10-5 N. in the + x direction.

AM
 
thank you for the help.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top