Question regarding Laplace's Equation for regions with charges

  • #1
Harikesh_33
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Why doesn't the **Laplace's equation**(#\nabla^2V=0#) hold in the region within the sphere when there is a charge inside it ? Is it because #ρ \ne 0# within the sphere and it becomes a **poisson equation**($\nabla^2V=\dfrac{-ρ}{ε_0}$) and changes the characteristics of **Harmonic Solution** (ie),

no maximum or minimum is allowed within the domain of the region in which the laplaces equation is used ? Now this condition becomes redundant because the PDE takes a non zero value .

Can we say that Laplace's equation takes that there is no net charge in the region ,so we add an additional factor of #\dfrac{Q_enc}{4πε_0R²}# to compensate the idea of no charge inside the sphere ?

 

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  • #2
Orodruin
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Is it because #ρ \ne 0# within the sphere and it becomes a **poisson equation**($\nabla^2V=\dfrac{-ρ}{ε_0}$)
Yes.
 
  • #3
Harikesh_33
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Yes.
Let me get this straight ,so Laplace's equation is used to find a potential function ,now if the potential obeys Laplace's equation then we can find the average potential around a sphere just by taking the potential at the centre of the sphere ,now here comes the catch ,Laplace's equation doesn't conside rthe charge at the centre ,so we add an additional Q_enc /4πε_0R to take the effect of charge at the centre into account .Am i right ?
 
  • #4
Orodruin
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Yes, although it would be more appropriate to say that the Laplace equation follows from the Poisson equation when you set the charge density to zero.
 
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  • #5
Harikesh_33
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Yes, although it would be more appropriate to say that the Laplace equation follows from the Poisson equation when you set the charge density to zero.
Yeah but won't the properties of the harmonic function vanish if we use Poisson's equation ?
 
  • #6
Harikesh_33
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Yeah but won't the properties of the harmonic function vanish if we use Poisson's equation ?
Yes, although it would be more appropriate to say that the Laplace equation follows from the Poisson equation when you set the charge density to zero.
I have a follow up question if you don't mind ,why is earnshaws theroem not true ? I understand that it has something to do with the fact that it's an unstable equilibrium(or it is said so and I don't understand why it is said so ) and potential is maximum for an unstable equilibrium and this isn't
 
  • #7
Orodruin
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What do you mean? Earnshaw's theorem says that there is no stable configuration for a set of point charges from electrostatic interactions alone. It is not really what you have been asking about here.

It is also not true that the potential is necessarily at a maximum for an unstable equilibrium. Saddle points are also unstable.
 
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  • #8
Orodruin
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Yeah but won't the properties of the harmonic function vanish if we use Poisson's equation ?
Yes, the solution is no longer harmonic in general - but it is harmonic in volumes where there is no charge since Poisson's equation then reduces to Laplace's equation.
 
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  • #9
Harikesh_33
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What do you mean? Earnshaw's theorem says that there is no stable configuration for a set of point charges from electrostatic interactions alone. It is not really what you have been asking about here.

It is also not true that the potential is necessarily at a maximum for an unstable equilibrium. Saddle points are also unstable.
Sorry for framing the question like that ,to provide some context ,if I am not mistaken ,Laplace's equation is being used to prove Earnshaw's theroem ,But how can we use Laplace's equation here ? Change density is non zero here .
 
  • #10
vanhees71
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Earnshaw's theorem is about properties of the electrostatic field in regions with no charge. A practical application is that you can't build a charged-particle trap with electrostatic fields alone. That's why you have, e.g., an additional magnetic field to construct a Penning trap.
 
  • #11
Harikesh_33
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Earnshaw's theorem is about properties of the electrostatic field in regions with no charge. A practical application is that you can't build a charged-particle trap with electrostatic fields alone. That's why you have, e.g., an additional magnetic field to construct a Penning trap.
Aren't we considering the charges placed on the centre and vertices of the cube ? And are you coming to the conclusion that it's a saddle point because both the first and second derivative is zero (follows Laplace's equation )
 
  • #12
Orodruin
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Aren't we considering the charges placed on the centre and vertices of the cube ? And are you coming to the conclusion that it's a saddle point because both the first and second derivative is zero (follows Laplace's equation )
No. The Earnshaw theorem would consider the regions where there are no charges and conclude that there is no stable point for a test charge in those regions. There is not a single second derivative (nor is there a single first derivative). The gradient being zero tells you that all the first derivatives are zero but the Laplace operator acting on the potential being zero tells you that the second derivatives in different directions must have different signs.
 
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