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A completely isolated neutral conducting sphere of radius $R$ is kept such that its center is at a distance of $r\left(>R\right)$ from a point charge $+Q$.

How can I find the force of interaction of the induced charges and the point charge, or at least the energy? I can't use "method of images" because the sphere is not grounded.

Note: The actual question has a sphere already charged with $+Q$ charge and it asks for the $r$ at which the point charge is in equilibrium. I thought of breaking the force down into a superposition of two forces, one from the $+Q$ of sphere and the other from the induced charges. If there is some other way to solve this, I would like to know that too.

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You can always solve the Laplace equation in spherical co-ordinates(using Legendre polynomials,etc).But even method of images will work.We have to use two image charges instead of one. – Sandesh Kalantre Jan 24 '14 at 12:36

(Ah, a very old question which is not yet answered... and I don't find a duplicate answer, either, though the problem is very nice ...
the comment is point you there, and that the user "centralcharge" has edited the question is amazingly fitting, too :) but it's not yet elaborated)

Yes, you can use the method of images. Normally you are right, it requires the object to be grounded, but for a sphere there is a nice "trick".

You have to first find an image, that would make the potential on the sphere zero. How to do this is apparently clear to you. This would solve the grounded case.
But there is no requirement for the potential to be zero. It has just to be equal on the sphere. And this requirement is not violated, if you put some charge in the center of the sphere. This central charge will just shift the potential on the sphere, but obviously not break spherical symmetry.
Take the extra charge just as big as you need to adjust the right net charge. Then you have solved the problem for non-grounded sphere.


So now to your full problem: In what distance would the sphere with $+Q$ net charge and the point charge of $+Q$ be at equillibrium?

You can solve this with an quite complicated equation or just guess the solution and check. Let's guess the golden ratio: $r = \frac{\sqrt 5 + 1}{2}R$, and briefly check this:

I'll set the radius of the sphere and the charges to $1$. Now assume the outside charge has distance $\phi = \frac{\sqrt 5 + 1}{2}$ from the center of the sphere. Then the induced charge is $-1/\phi$ and it's distance from the center is $1/\phi$ too. The charge in the middle has to be $1+1/\phi$ for the net charge to be $1$
Now it holds $1/\phi = \phi-1$, that's why $\phi$ is nice. It follows:
The force between the outside charge and the induced charge: the distance is $1$, the product of charges is $1/\phi$, so the force is an attraction of $\frac{1/\phi}{1^2} = 1/\phi$.
The force between the outside charge and the central charge: the distance is $\phi$, the product of charges is $\phi$, so the force is a repulsion of $\frac\phi{\phi^2} = 1/\phi$.
The forces are equal, qed.

(...if you solve the equations you see, that this is indeed the only real solution greater than 1)

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