Force on 3uC Charge - Solving Coloumb's Law Problem

  • Thread starter Thread starter Ember Cult
  • Start date Start date
  • Tags Tags
    Coloumbs law Law
AI Thread Summary
To find the force on a 3uC charge at point A due to other charges at points B and C, the forces must be calculated as vectors. The total force FA is the vector sum of the forces from each charge, represented as FA = FAB + FAC. Each force should be calculated using Coulomb's Law, ensuring to account for the distance between charges, as the force magnitude is inversely proportional to the square of that distance. It's crucial to consider the direction of each force and resolve them into horizontal and vertical components before summing. Properly applying these principles will yield the correct net force on the 3uC charge.
Ember Cult
Messages
2
Reaction score
0

Homework Statement


Find force on 3uC charge.
7c3691e89fc776fddf6f46456d2b01b3.png


Homework Equations


3f67479eda67aa1df0b2b44ca10a116d.png


The Attempt at a Solution


Assuming the 3uC charge is at point A and others are at point B and C

FA = FAB + FAC
FAB = K*(2*3)/5
FAC = K*(2*3)/5

 
Physics news on Phys.org
Ember Cult said:

Homework Statement


Find force on 3uC charge.
7c3691e89fc776fddf6f46456d2b01b3.png


Homework Equations


3f67479eda67aa1df0b2b44ca10a116d.png


The Attempt at a Solution


Assuming the 3uC charge is at point A and others are at point B and C

FA = FAB + FAC
FAB = K*(2*3)/5
FAC = K*(2*3)/5

Remember, force is vector. The two forces exerted on the 3uC charge have different directions. Add them by their horizontal and vertical components.
The magnitude of the Coulomb force is inversely proportional to the square of the distance. Correct your formulas.
 
ehild said:
Remember, force is vector. The two forces exerted on the 3uC charge have different directions. Add them by their horizontal and vertical components.
The magnitude of the Coulomb force is inversely proportional to the square of the distance. Correct your formulas.
Your relevant equation should include unit vector of r. Calculating 2 charge particles is not a problem but not in 3D or many particles in 2D.
 
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