Constant potential inside spherical shell

AI Thread Summary
In a spherical shell made of a perfect conductor, the electric field inside is zero, leading to a constant potential. When a charge is placed at the center of this shell, it experiences no net electric force due to the constant potential. If the charge is slightly displaced, it will move in the direction it was pushed at a constant velocity, assuming no other forces act on it. This behavior aligns with the principle that a constant potential indicates no electric field, and thus no force is exerted on the charge. Understanding this concept clarifies the relationship between electric potential and force in electrostatics.
physicsjock
Messages
84
Reaction score
0
Hey,

I just wanted to double check if what I am thinking is correct.

Say you have a spherical shell of inner radius R1, and outer radius R2, which is made of a perfect conductor carrying a charge q1.

E=0 inside (r<R1) (and also between R1<r<R2 but not worried about that)

So the potential inside the shell is constant.

Now, say there is another charge centered inside the shell,

Is it correct to think that the charge remains in its position as the potential inside the shell is constant?

So if you were to bump the charge slightly (very slightly so its barely moving) it would simply move in the direction it was pushed at a constant velocity? And this is because the potential is constant within the shell so there is no other force acting on the charge?


Thanks in advance,
 
Physics news on Phys.org
physicsjock said:
Hey,

I just wanted to double check if what I am thinking is correct.

Say you have a spherical shell of inner radius R1, and outer radius R2, which is made of a perfect conductor carrying a charge q1.

E=0 inside (r<R1) (and also between R1<r<R2 but not worried about that)

So the potential inside the shell is constant.

Now, say there is another charge centered inside the shell,

Is it correct to think that the charge remains in its position as the potential inside the shell is constant?

So if you were to bump the charge slightly (very slightly so its barely moving) it would simply move in the direction it was pushed at a constant velocity? And this is because the potential is constant within the shell so there is no other force acting on the charge?
Well, there's no net electric force acting on the charge; but there's no evidence about any other forces that may or may not be acting (gravitational, inertial).
 
Yea fair enough, I didn't mention any other forces.

I was just trying to understand constant potential a little better.

Disregarding any other outside forces, is it correct to say since the potential is constant (no force acting), if the charge inside the spherical shell were pushed it would simply move in the direction it was pushed at a constant velocity (assuming it was only pushed a very small amount)?
 
physicsjock said:
Yea fair enough, I didn't mention any other forces.

I was just trying to understand constant potential a little better.

Disregarding any other outside forces, is it correct to say since the potential is constant (no force acting), if the charge inside the spherical shell were pushed it would simply move in the direction it was pushed at a constant velocity (assuming it was only pushed a very small amount)?
Yes, it is true for "mundane" scenarios (non relativistic velocities).

I'll just point out that a constant potential means a zero electric field, since the field is the gradient of the potential. No field means no force.
 
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