Where can a third charge +Q be placed so that it experiences a zero force?

AI Thread Summary
To achieve zero force on a third charge +Q placed between a charge of +2q at the origin and a charge of -q at x=3.0 cm, it cannot be positioned in the middle, as it would experience a net force directed to the right. The third charge must be placed either to the left of +2q or to the right of -q. The electric field generated by the two charges influences the force acting on +Q, and the relevant equation for force is F = KQQ/r². Understanding the balance of forces is crucial for determining the correct placement of charge +Q. Proper visualization and application of electric field concepts are essential to solve this problem.
shashaeee
Messages
30
Reaction score
0
I'm having a hard time trying to picture the problem:

A charge of +2q is placed at the origin and a second charge of -q is placed at x= 3.0cm. Where can a third charge +Q be placed so that it experiences a zero force?

Here is how I picture the problem:

physics4.jpg
 
Physics news on Phys.org
shashaeee said:
I'm having a hard time trying to picture the problem:

A charge of +2q is placed at the origin and a second charge of -q is placed at x= 3.0cm. Where can a third charge +Q be placed so that it experiences a zero force?

Here is how I picture the problem:

View attachment 49498

The 3rd charge could end up anywhere on the line, left, right or between the first 2 charges. What is the equation for the electric field E generated by a charge at some distance...?
 
For Force its, F = KQQ/r2

The way I understand the problem is it could either be on the left or right of those two charges. Because if it was in the middle, the third charge will experience a net force directed to the right. Right?
 
shashaeee said:
For Force its, F = KQQ/r2

The way I understand the problem is it could either be on the left or right of those two charges. Because if it was in the middle, the third charge will experience a net force directed to the right. Right?

Yes, that is correct. It's good that you see that.
 
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