How many external forces act on each dipole and why?

AI Thread Summary
Each dipole experiences two external forces due to the charges of the adjacent dipole. Specifically, dipole A feels the forces from the positive and negative charges of dipole B. The total number of external forces acting on each dipole is two, not four, as previously assumed. It is crucial to recognize that forces acting on different dipoles should not be combined. Thus, the correct understanding is that each dipole is influenced solely by the charges of the other dipole.
Lee33
Messages
156
Reaction score
0

Homework Statement



Two identical permanent dipoles, each consisting of charges ##+q## and ##-q## separated by a distance ##s##, are aligned along the ##x## axis, a distance ##r## from each other, where ##r >> s.##

a. How many external forces act on each dipole?

Homework Equations



Columb's law

The Attempt at a Solution



For a, I got for each dipole only two external forces acts on them from the dipole next to them. So for the net external forces it will be four. But its wrong, why is that?
 
Physics news on Phys.org
How many external forces act on each dipole?

You have dipole A and dipole B. What external forces act on dipole A?

ehild
 
Dipole B will be an external force on A.
 
Dipole B is a dipole, not a force. And you wrote previously that it acts by two forces on dipole A. What are they?
B exerts two forces on dipole A, and also dipole A exerts two forces on dipole B. The question was the number of forces acting on one dipole, and it is the same for both dipoles. You can not add forces acting on different bodies.
ehild
 
So, the external force acting on dipole A will be the only the two charges making up dipole B?
 
Lee33 said:
So, the external force acting on dipole A will be the only the two charges making up dipole B?

The only external forces on dipole A will be the forces exerted on both its charges by the two charges, making up dipole B.
One dipole has its electric field http://en.wikibooks.org/wiki/Physics_Exercises/Electrostatics and you can determine the force on both charges of the other dipole in that field.

ehild
 
  • Like
Likes 1 person
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