Charge inside a charged sphere (Gauss law)

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Suppose we have a sphere with a uniform surface charge. According to Gauss law the electric field inside such a sphere is zero. So if I put a charge ( let me call it Q from here) in it, it wouldn't experience any force, would it?

What I don't understand is this: the charge Q we actually put inside does create an electric field so our sphere should move thus it only comes down to picking the right point of reference so one could say that it is our charge Q which is actually moving

However, if the sphere can't move and I put my charge not in the center of the sphere it would feel a stronger attraction to the closest point than to any other thus should move and if it moves there is an electric field inside.

Thus according to my logic in both cases the charge inside is affected :confused:

Please help me out, I am really struggling on this one
 

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  • #2
olgranpappy
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Suppose we have a sphere with a uniform surface charge. According to Gauss law the electric field inside such a sphere is zero. So if I put a charge ( let me call it Q from here) in it, it wouldn't experience any force, would it?

What I don't understand is this: the charge Q we actually put inside does create an electric field so our sphere should move thus it only comes down to picking the right point of reference so one could say that it is our charge Q which is actually moving

However, if the sphere can't move and I put my charge not in the center of the sphere it would feel a stronger attraction to the closest point than to any other thus should move
No. This is wrong. This can either be proved explictly or by appealing to gauss' law and symmetry (there still *is* a symmetry because we are calculating the field due to the shell not the charge Q since the charge Q doesn't act on itself). I.e., as you already said gauss's shows that E=0 inside the spherical shell. Thus the force on Q is zero and thus there is no *net* attraction to the closest point or any other point for that matter. Because of the 1/r^2 nature of the force the farther points (of which there are more) can exactly conspire to cancel the forces from the "closer" points of which there are less.

and if it moves there is an electric field inside.
true but it does not move and thus there is no electric field as stated at the begining.

Thus according to my logic in both cases the charge inside is affected :confused:

Please help me out, I am really struggling on this one
 
  • #3
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Thanks

I realise that from Gauss law I do get E=0 which means there is no force on a charge inside the sphere, however suppose we didn't know Gauss law. For me it's just hard to grasp it conceptually, although this

Because of the 1/r^2 nature of the force the farther points (of which there are more) can exactly conspire to cancel the forces from the "closer" points of which there are less.
seems like a good kick in the right direction
 
  • #4
olgranpappy
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Thanks

I realise that from Gauss law I do get E=0 which means there is no force on a charge inside the sphere, however suppose we didn't know Gauss law. For me it's just hard to grasp it conceptually, although this



seems like a good kick in the right direction
if you didn't know gauss' law but you still knew the coulomb force law
[tex]
d\bold{F}=\frac{k Q dq}{|\bold{r}-\bold{R}|^3}(\bold{r-R})\;,
[/tex]
where R is the location of Q and r is a point on the spherical surface (of radius L and surface charge
density [itex]\sigma[/itex] giving [itex]dq=L^2d\Omega\sigma[/itex] where [itex]d\Omega[/itex] is the solid angle [itex]d\Omega=d\phi \sin(\theta)d\theta[/itex]),
you can certainly work out the net force explicitly. We have
[tex]
\bold{F}
=
kQ\sigma L^2 \int d\Omega \frac{\bold{r-R}}{\left|{\bold{r}-\bold{R}}\right|^{3}}\;.
[/tex]

Without loss of generality we can choose R along the z-hat direction of the variable r.

Then F_x and F_y are easily seen to be zero.

And the only remaining thing that we must show equals zero (the hard part--the important part) is
F_z which is proportional to the integral
[tex]
\int_{-1}^{1}dx\frac{(Lx-R)}{(L^2+R^2-2LRx)^{3/2}}\;,
[/tex]
which (I can assure you:wink:) is zero if L>R and is -2/R^2 if R>L.

But since the charge Q is inside the sphere L>R and so F_z=0.

If you don't have faith in my assurance that the above integral is zero (which you should not because you really have no reason to trust me:wink:) you can work it out for yourself. Cheers.
 
  • #5
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I never worked with the solid angle. I'll look into it and return to your post as soon as I figure it out, thanks
 
  • #6
olgranpappy
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okay. Also, I think this calculation is in a lot of upper level undergraduate physics textbooks. You don't necessarily have to look in electromagnetism textbooks either. For example, there is a completely analogous calculation for the gravitational case which can probably be found in a lot of classical mechanics texts. Good luck.
 

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