How Does the Method of Images Determine Potential Inside a Grounded Sphere?

[crimsonidol]

Homework Statement

We have a grounded conducting sphere of radius R and a charge q is placed a distance a from the center of the sphere. Show that the potential in the interior produced by q and the distributed induced charge is the same as that produced by q and its image charge q'.The image charge is at a distance a'=R^2/a fro the center collinear with q and the origin. Calculate the electrostatic potential for a<R<a'. Show that potential vanishes for r=R if we take q'=-qR/a

Homework Equations

In the course we are investigating legendre equations, legendre polynomials etc.

The Attempt at a Solution

I can find potential by simply using V=Sum (1/4pieps)q/r however there is no legendre polynomial or legendre series in it. I tried Laplace's Equation however I got confused. Because in Laplace's eqn I can only deal with r. Where have I gone wrong? How should i think?

 

[crimsonidol]

I just need a spark?

 

[ehild]

Read this. http://en.wikipedia.org/wiki/Method_of_image_chargesehild

 

[crimsonidol]

I know that solution, however there is nothing related to legendre in it. The question was given to me as a homework for Legendre equation/polynomials topic. I should o something related to that I suppose.

 

[rwisz]

One way to approach this problem is to use the method of images. This technique allows us to simplify the problem by introducing an imaginary charge (image charge) at a specific location that produces the same potential as the actual charge in the given configuration.

To start, let's consider the potential at a point P inside the conducting sphere, created by the real charge q at a distance a from the center. We can write this potential as:

V = (1/4πε₀) * (q/r + q'/r') (1)

Where r is the distance between the point P and the real charge q, and r' is the distance between the point P and the image charge q'.

Now, to find the value of q', we need to consider the boundary condition at the surface of the conducting sphere, where the potential must be zero. This means that when r=R, the potential must be zero. Substituting this in equation (1), we get:

0 = (1/4πε₀) * (q/R + q'/R') (2)

Now, we know that the image charge is located at a distance a' from the center of the sphere, collinear with the real charge q and the origin. This means that a' = R²/a. Substituting this in equation (2), we get:

0 = (1/4πε₀) * (q/R + q'/(R²/a)) (3)

Solving for q', we get:

q' = -qR/a (4)

Now, substituting this value of q' in equation (1), we get the potential at point P inside the conducting sphere as:

V = (1/4πε₀) * (q/r - qR/a*r') (5)

Now, we can use this potential to find the potential at any point inside the conducting sphere, by simply considering the distance r from the center of the sphere to the point P. This potential will be the same as the potential produced by the real charge q and the distributed induced charge on the surface of the conducting sphere.

In conclusion, by using the method of images, we have shown that the potential inside the conducting sphere produced by the real charge q and the distributed induced charge is equivalent to the potential produced by the real charge q and its image charge q'. This method allows us to simplify the problem and use known equations to find the potential