Principle behind electrostatic shielding?

AI Thread Summary
The electric field inside a solid conducting sphere is zero due to the equilibrium of electrons, as any net force would disrupt this balance. In a hollow conducting sphere, the electric field remains zero because the interior is surrounded by an equipotential shell, ensuring uniform potential throughout. Even when external charges are present, they only affect the charge distribution on the outer surface, leaving the interior field unaffected. The principle is similar to gravitational effects, where only mass within a certain radius influences the gravitational field at that point. Thus, the absence of internal sources leads to no electric field inside the hollow sphere.
Dr.azwar
Messages
10
Reaction score
0
if we have a solid conducting sphere with charges around it, then the elctric field inside the sphere is zero otherwise the electrons of the sphere would not be in equilibrium as there would be a net force acting on it. however if its a hollow sphere then why does the electric field inside the hollow sphere be zero?
 
Physics news on Phys.org
Because the inside of the sphere is conducting, the potential is the same everywhere. There are no other sources of field or potential inside. No potential difference ##\Rightarrow## no electric field. Simple, isn't it ?
 
what if the hollow sphere is surrounded by charges, the field is still zero.how is that possible?(the empty space inside the hollow sphere is not conducting)
 
The empty space inside the hollow sphere is surrounded by an equipotential shell: the inside of the conducting sphere. No sources inside so the entire hollow sphere is at the same potential (*). Thus: no field.

Surrounding the conducting sphere with charges only causes an uneven charge distribution on the outside surface of the conducting sphere. Again, on that outside surface the potential has to be the same everywhere (otherwise the charges would simply move until it's the same). But on the outside there is a contribution from those external sources (charges).

(*) Note that that potential does not have to be zero: e.g. a lot of positive charges on the outside means that the inside surface charge is negative.

--
 
Dr.azwar said:
if we have a solid conducting sphere with charges around it, then the elctric field inside the sphere is zero otherwise the electrons of the sphere would not be in equilibrium as there would be a net force acting on it. however if its a hollow sphere then why does the electric field inside the hollow sphere be zero?
The vector sum of the E field due to all the charges is zero inside the shell.

Same reason that, if you dig a hole into the Earth to a radius r' < R where R is the radius of the Earth, gravity at r' is due to the mass INSIDE r' only; mass at r > r' is - pardon the pun - immaterial.
 
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