Charge at Equilibrium Exercise, .

AI Thread Summary
When two conductor spheres with radii R and 2R are connected, the charge will redistribute based on their surface areas, leading to an equilibrium state. The smaller sphere with charge Q=+1.7C will have a charge of +0.57C, while the larger sphere will have +1.13C. If the spheres were not at a great distance, the electric fields from each sphere would influence the charge distribution, complicating the equilibrium calculation. The principle of conservation of charge ensures that the total charge remains constant during redistribution. Understanding the relationship between charge density and surface area is crucial for determining the final charge on each sphere.
claudiadeluca
Messages
9
Reaction score
0
Please help with the following exercise:

Two conductor spheres with radius R and 2R are at a great distance one from the other. We put a charge of Q=+1.7C on the smaller sphere, and then we put them in contact with a conductor thread. Determine what charge the spheres will have at equilibrium. What would not work in our reasoning if the two spheres weren't at a great distance one from the other?

I don't know what relation there is between what we are trying to find and the radius of the sphere. I know that if the spheres had the same radius, the charge at equilibrium between the two would be +Q/2.
I don't know what would not work if the spheres were close, maybe the electric field influencing the motion of charge? I have no clue, please help.

Thank you.
 
Physics news on Phys.org
What would not work in our reasoning if the two spheres weren't at a great distance one from the other?
That would infer that the electric field caused by the electric charge on either sphere does not appreciably affect the other charge on the other sphere. Think of an electric field, which must originate from a charge somewhere.

Anyway, there is a conservation of charge!

The charge starts on one sphere, and has a specific charge density.

When the two spheres are connected by a conductor, the charge redistributes such that the charge density must be equal.

So for the given charge, determine the apportionment of the total charge assuming the same charge density and surface area of each sphere.
 
Thank you very much, your help was precious.
 
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