Electostatic(Guass' Law)

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In summary: the fieldlines are already there and they will keep moving in the same direction because there is no net electric field present.
  • #1
kimchi
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a metal spherical shell with a +q in it but not centered (in the cavity), why the induced +ve on outer surface is uniformly distributed? (the shell is neutral, and the induce -ve and +ve charges not shown.)
 
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  • #2
kimchi said:
a metal spherical shell with a +q in it but not centered (in the cavity), why the induced +ve on outer surface is uniformly distributed? (the shell is neutral, and the induce -ve and +ve charges not shown.)

Very good question. This is a classic. the clue is that the shell must remain electrically neutral. This implies that no net-electric fields can be generated inside the shell (inside the material that is) and then the flux through the gaussian surface must be ZERO inside the shell. You will only have an electric field due to +q , enclosed by the shell and an electric field outside the shell.

Now suppose you draw a Gaussian surface inside the material of the shell. Since the enclosed charge is +q within the shell, there will have to be a charge -q on the inner walls of the shell (that part that is still enclosed by the Gaussian surface). the distribution of these charges will not be uniform because q is not at the center.

But, since the shell must be neutral and now it's inner wall has charge -q, there will have to be a charge +q that goes from the inner wall to the outside of the shell (that part of the material that is outside the gaussian surface : you see, then the flux through the gaussian surface is indeed zero). The charges will spread out UNIFORMLY because there won't be an netto-electric field inside the shell's material, that is able to deflect these positive charges as they go outside the Gaussian surface. They all go from the inner wall to the outer wall in the exact same way : uniform distribution

marlon
 
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  • #3
my interpretation is that...for electrostatic equilibrium
the electric field of any pt in a conductor is 0. Inside a G. surface as the outer surface of the shell, electric flux is 0. Image there are no charge inside as induced -ve= +, the induced + charge must be uniformly spreaded outside the shell so as to keep net E-field inside the shell 0 (not include the cavity)

but i don't understand: from Gauss' Law>>flux is 0 inside the Gaussian surface when 0 charge enclosed in it, but should E-field be 0 too(inside the shell material)??
 
  • #4
kimchi said:
my interpretation is that...for electrostatic equilibrium
the electric field of any pt in a conductor is 0. Inside a G. surface as the outer surface of the shell, electric flux is 0. Image there are no charge inside as induced -ve= +, the induced + charge must be uniformly spreaded outside the shell so as to keep net E-field inside the shell 0 (not include the cavity)

but i don't understand: from Gauss' Law>>flux is 0 inside the Gaussian surface when 0 charge enclosed in it, but should E-field be 0 too(inside the shell material)??

Electrostatic equilibrium means that there is no NETTO-electric field. This means that through the gaussian surface there will flow an equal amount of positive and negative charges in opposite direction. Both flows (each being non-zero) cancel out each other. So the net electric field : E_N = E - E = 0
This is what happens in this case. Indeed in a conductor, the E-field is always zero due to the above reason.

When the gaussian surface encloses a zero charge, then everything needs to be zero. Just apply Gauss's law when q = 0...That is quite starightforward, isn't it ?

Inside the material the total electric field is zero because the outer shell has E and the inner shell has -E, so if you add up these two you get zero. ofcourse locally (ie in each part of the shell) there is a non-zero electric field and that is why we will have fieldlines.

regards
marlon
 
  • #5
Kimchi,

Just as an addendum : you cannot say that a flux inside the Gaussian surface is ZERO. It's definition really states that you can only speak about a flux that passes through some surface. Flux is a measure for the number of fieldlines that pass through a surface (perpendicular to it)

marlon
 
  • #6
but i ain't sure why should + Q evenly distributed outside the gaussian surface? as what i said above? for ensure the 0 E-field!
 
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  • #7
kimchi said:
but i ain't sure why should + Q evenly distributed outside the gaussian surface? as what i said above? for ensure the 0 E-field!

But that is what i explained in my very first post. +Q is uniformely distributed in the outer region of the shell because there is no electric field present in this region, PRIOR to when the +Q charges run off to the outer region. So when these charges 'enter' the outer region (ie as the have passed through the Gaussian surface) they are in a region with no present electric field. The electric field of the outer region (ie the field lines you get outside the shell) is a result of these uniformely distributed charges. But, like i said, that is just the E-field of these charges and there was no E-field acting on these charges when they entered the outer region. No electric field means no force on these charges and thus no deflection in their paths

marlon
 
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  • #8
i guess i know all such thing...but the inner induced -ve charges are not evenly distributed...while outer induced +ve charges are evenly distributed, right? thanks very much
 
  • #9
kimchi said:
i guess i know all such thing...but the inner induced -ve charges are not evenly distributed...while outer induced +ve charges are evenly distributed, right?

YES INDEED

this was your very question, remember ?

regards
marlon

ps : now, i ask you : WHY IS THAT ?
 
  • #10
-ve induced Q more concentrated at inner surface near the biased + Q in the cavity, any G. surfaces in the shell material enclosed no charges inside, so no flux via them.
as elctrostatic equilibium, any conductor is in E=0, so the outer induced +ve Q must evenly distributed to account again no electric field in the shell.
 
  • #11
kimchi said:
- G. surfaces in the shell material enclosed no charges inside, so no flux via them.

Incorrect, a G surface inside the shell ofcourse encloses charges : first of all the +Q charge that is at the inside of the shell. The flux through this surface must be ZERO because the E-field in the shell must be zero. But because of the +Q charge inside the shell, negatice charges will move towards the inner region of the shell. this means they pass through the G.surface yielding a non-zero flux. In order to compensate for that positive charges will move through the G surface in the opposite direction. Then the net flux is ZERO.

The negative charges will move towards the E-field of the inside +Q charge, yielding the non-uniform distribution. The positive charges in the actual shell will move outward to a region were there is no E-FIELD (because the E-field inside the shell-material is zero) yielding the uniform distribution

marlon
 
  • #12
more clear now. thanks
my mistake is focus on result=final state.
but process is essential too.
 
  • #13
kimchi said:
more clear now. thanks
my mistake is focus on result=final state.
but process is essential too.

sure it is

marlon
 

What is Gauss' Law?

Gauss' Law is a fundamental law of electromagnetism that describes the relationship between electric charges and electric fields. It states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.

How is Gauss' Law derived?

Gauss' Law is derived from the principles of Coulomb's Law and the superposition principle. By considering a small test charge placed at different points on a closed surface, we can determine the total electric flux through that surface and ultimately derive the equation for Gauss' Law.

What is the significance of Gauss' Law?

Gauss' Law is significant because it allows us to calculate electric fields in situations with high degrees of symmetry, where direct application of Coulomb's Law may be difficult. It also provides a way to calculate the electric field inside a conductor or on a charged surface.

Can Gauss' Law be applied to all situations?

No, Gauss' Law can only be applied to situations with high degrees of symmetry, such as a point charge, line charge, or charged sphere. For more complex situations, other methods such as numerical integration or the method of images may be used.

How is Gauss' Law related to electric potential?

Gauss' Law is closely related to the concept of electric potential. By using the gradient of the electric potential, we can derive the electric field from the electric potential, and vice versa. This relationship is known as the gradient theorem and is an important tool in solving electrostatic problems.

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