Electric Charges Causing Centripetal Motion

AI Thread Summary
A particle of charge Q is fixed at the origin, while another particle with mass 0.671 g and charge 5.90 µC moves in a circular path due to the electric force exerted by Q. The relevant equations include Coulomb's law and centripetal force, leading to the equation kqQ/r^2 = mv^2/r. After calculations, it was determined that Q should equal approximately 5.8 μC for the moving particle to maintain circular motion. However, the solution was marked incorrect, prompting a discussion about potential errors in the calculations and the nature of the force involved. The direction of the force, whether attractive or repulsive, is also highlighted as a critical factor in solving the problem.
mdf730
Messages
5
Reaction score
0

Homework Statement



A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.671 g, q = 5.90 µC is located on the x-axis at x = 15.6 cm, moving with a speed of 54.4 m/s in the positive y direction. For what value of Q (in μC) will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)

Homework Equations



F=F
F=kqQ/r^2
F=mv^2/r

The Attempt at a Solution



F=F
kqQ/r^2=m(v^2)/r
kqQ=m(v^2)r
(8.99*10^9)(5.9*10^-6 C)(Q) = (.000671 kg)((54.4 m/s)^2)(.156 m)
53100 * Q = .30977
Q= .0000058337 C
Q= 5.8 μC

I have not been able to find my error, but I did something wrong, as the wileyplus marks it incorrect.
 
Physics news on Phys.org
mdf730 said:

Homework Statement



A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.671 g, q = 5.90 µC is located on the x-axis at x = 15.6 cm, moving with a speed of 54.4 m/s in the positive y direction. For what value of Q (in μC) will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)

Homework Equations



F=F
F=kqQ/r^2
F=mv^2/r

The Attempt at a Solution



F=F
kqQ/r^2=m(v^2)/r
kqQ=m(v^2)r
(8.99*10^9)(5.9*10^-6 C)(Q) = (.000671 kg)((54.4 m/s)^2)(.156 m)
53100 * Q = .30977
Q= .0000058337 C
Q= 5.8 μC

I have not been able to find my error, but I did something wrong, as the wileyplus marks it incorrect.
Welcome to PF!

What is the direction of the force? (Hint: attractive or repulsive?).

AM
 
Ahh. Thank you!
 
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 '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