A point charge in a sphere with V0

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SUMMARY

This discussion focuses on solving the voltage and electric field distribution for a point charge located at the center of a hollow sphere with a voltage of V0. The key approach involves using the non-homogeneous partial differential equation (PDE) represented by ∇²V = q/ε₀, where q is the charge inside the sphere. The spherical symmetry simplifies the problem, allowing the application of Gauss' Law and the capacitance of the sphere to determine the electric field and potential distributions. The solution requires combining particular and homogeneous solutions, potentially utilizing eigenfunction expansion and Legendre polynomials.

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rootX
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If there is a point charge at the center of a hollow sphere that has voltage of V0. How would I go by solving the voltage and electric field distribution?

I know I can use Poisson (Laplace) to find distribution had there been no charge inside the sphere, and could use Gauss had there been no V on the sphere. But how about the combination?
 
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Hmmm. So there is a spherical shell with a voltage, and now we want to find the voltage inside the shell?

So it seems that what you'll have to do, unless I'm overlooking something, is

\nabla^2 V = \frac{q}{\epsilon_0}

You'll be solving a non-homogeneous PDE. It may be a tricky solution to work out. Of course, once you solve V, you can E from the gradient (which also may be not easy task).
 
The spherical symmetry makes this into a simple problem.

Since the point charge is placed at the very center of the sphere, that means that whatever charge redistribution it causes on the hollow sphere, retains the spherical symmetry. That means that the sphere provides no inwards E-field and acts as a point charge outwards (Remember that no charge has flowed off the sphere or onto it).

Use Gauss' Law and the capacitance of the sphere to solve for the charges, potentials and fields.
 
Mindscrape said:
Hmmm. So there is a spherical shell with a voltage, and now we want to find the voltage inside the shell?

So it seems that what you'll have to do, unless I'm overlooking something, is

\nabla^2 V = \frac{q}{\epsilon_0}

You'll be solving a non-homogeneous PDE. It may be a tricky solution to work out. Of course, once you solve V, you can E from the gradient (which also may be not easy task).

There is a spherical shell with a point charge Q at its center and I want to know how to find E/V distribution over all the space inside/outside.

\nabla^2 V = \frac{q}{\epsilon_0}

So the "q" will be just the charge that is located inside the sphere?
RoyalCat said:
The spherical symmetry makes this into a simple problem.

Since the point charge is placed at the very center of the sphere, that means that whatever charge redistribution it causes on the hollow sphere, retains the spherical symmetry. That means that the sphere provides no inwards E-field and acts as a point charge outwards (Remember that no charge has flowed off the sphere or onto it).

Use Gauss' Law and the capacitance of the sphere to solve for the charges, potentials and fields.

Because, there is a boundary condition that V0 is placed on the shell doesn't that affect the way Q induces charge over the shell or it can be just
Q/A
 
rootX said:
There is a spherical shell with a point charge Q at its center and I want to know how to find E/V distribution over all the space inside/outside.

\nabla^2 V = \frac{q}{\epsilon_0}

So the "q" will be just the charge that is located inside the sphere?

Right, this is just verbatim of Maxwell's Equation. It's been a while since I've solved a non-homogenous PDE, but it's just like for differential equations: the solution is the particular solution plus the homogeneous solution (the general solution is a well-known solution for this type of problem). Now generally, you'd use eigenfunction expansion to solve this, but I'm not sure how well that will work with Legendre polynomials. However, you should be able to just guess the particular solution. After all, you only need to figure out something that will give you a constant after two derivatives.

Again, I haven't yet worked it out myself, but maybe it won't be so bad now that I think about it. Let me know how it goes (or went if this is already past your due date).
 

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