Repulsion Forces Between Northern & Southern Hemisphere of Metal Sphere

  • Thread starter Thread starter Lea
  • Start date Start date
  • Tags Tags
    Forces Repulsion
AI Thread Summary
The discussion centers on the forces of repulsion between the northern and southern hemispheres of a charged metal sphere. It is established that while the electric field inside the sphere is zero due to charge distribution, the charges on the surface still exert forces on each other. The charges create a net radial electric field that acts outward, leading to mutual repulsion between the hemispheres. This effect occurs because, although the fields cancel within the sphere, they do not cancel at the surface where the charges reside. The conclusion reached is that the hemispheres do indeed repel each other due to the outward electric field acting on the surface charges.
Lea
Messages
3
Reaction score
0

Homework Statement


A metal sphere of radius R carries a total charge Q. What is the force
of repulsion between the "northern" hemisphere and the "southern" hemisphere?

Homework Equations

The Attempt at a Solution


Since the sphere is metallic, its conducting and so the electric field inside the sphere is 0, and outside the sphere it is perpendicular to the surface and is radially outwards, and so, the force exerted by the northern hemisphere on the southern hemisphere should be 0. However, the answer appears to be other than zero. why?
 
Physics news on Phys.org
The net field inside is zero because the all the individual fields from the individual charges distributed around the sphere cancel each other out in the interior. That does not say that those same fields don't serve to push all the charges to the skin of the sphere, keeping them spread out evenly over the surface.

As far as the mutual repulsion of hemispheres goes, you've still got two charge distributions pushing on each other.
 
But those charge distributions have an electric field that is radially outwards, how can they push on each other?
 
Lea said:
But those charge distributions have an electric field that is radially outwards, how can they push on each other?
Each charge has a spherically symmetric field radiating outward around itself. Exterior to the sphere the net effect of summing all the fields leaves just the radially outward part. Within the sphere they all cancel. The overall effect is to be able to perceive no field in the interior of the sphere and a radially outward field exterior to the sphere.

At the surface where the charges reside, they are feeling the effects of the net radial field. This is what keeps those charges confined to the outer surface. The fields aren't mutually canceled from there outward, so each of the individual charges is "feeling" a net field that looks as if all the charge Q were placed as a point charge at the center of the sphere (and the conducting sphere is ignored).
 
Thank you! I understand now.
 
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