Potential inside an non conducting sphere given the potential on the sphere.

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Given the equation for the potential "on" a non conducting sphere how can the potential and electric field inside the sphere be calculated?
If there is a potential on the sphere am I to assume that there must therefore be a charge build up in the sphere? Is this just calculated by the Laplace of the potential? How does the potential inside the sphere differ from that on the sphere?
 

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  • #2
K^2
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If you do not know the charge distribution inside the sphere, you cannot solve this problem. I would guess that what you are actually looking for is field inside a sphere of linear dielectric with surface charge distribution and neutral interior. In that case, yes, you solve Laplace equation. You know that [itex]\nabla \cdot E = 0[/tex] from Maxwell's eqns, and [itex]E = -\nabla V[/itex] is the equation for potential. In static case, you get [itex]\nabla^2F = 0[/itex], which you solve for given boundary conditions. If you decompose your surface potential into spherical Bessel functions, it should be a snap.
 
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Thank you for your reply.

All I am given is:
The potential on a non-conducting sphere or radius a is given by
V = V0(cos2[tex]\theta[/tex] +cos[tex]\theta-1)[/tex]

And with that I should be able to find the potential and electric fields inside and outside the sphere and the surface charge density?
 
  • #4
Born2bwire
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Thank you for your reply.

All I am given is:
The potential on a non-conducting sphere or radius a is given by
V = V0(cos2[tex]\theta[/tex] +cos[tex]\theta-1)[/tex]

And with that I should be able to find the potential and electric fields inside and outside the sphere and the surface charge density?
K^2 is slightly mistaken since the divergence of the electric field is related to charge, we can express charge in terms of the Laplacian operator on the potential. Specifically,

[tex] \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon} [/tex]

[tex] \mathbf{E} = -\nabla V [/tex]

[tex] \nabla^2 V = -\frac{\rho}{\epsilon} [/tex]

The applied electric field will induce a charge distribution which is why the right-hand side of Gauss' Law is non-zero here even though the original sphere has no net charge.

Since you are already given the potential distribution on the surface, it is simple vector calculus to find the applied field and induced charge distribution on the surface.

Now finding the distribution of charge, the potential, and the field INSIDE the sphere is a different matter as K^2 is talking about. You then have to solve for the Poisson Equation. First, the solution to the Laplacian Equation,

[tex] \nabla^2 V = 0 [/tex]

gives you your homogeneous solution. Then you have to solve for the Poisson Equation given the boundary condition for the voltage and charge on the surface.

[tex] \nabla^2 V = -\frac{\rho}{\epsilon} [/tex]

This is usually done by decomposing the solution and boundary condition into spherical modes and solving for the unknown coefficients by mode matching with the boundary condition.

EDIT: Griffiths' book gives a good explaination on how to do this.
 
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