Understanding Image Charge and Potential in Electrostatics

In summary, the questioner is looking for a solution to a problem involving potential. They ask for a more difficult way to do it than the first solution, and you provide one.
  • #1
crimsonidol
29
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1. 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


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


3. 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?
 
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  • #2
i just need a spark :(
 
  • #3
I'm not sure exactly what the questioner is expecting from you for this problem. The easy way to show that the actual potential is the same as the potential you get from the method of images is to simply show that they satisfy the same boundary conditions and differential equation (Laplace's equation) in the given region (interior of sphere) and then appeal to the uniqueness theorem. Another, more difficult way is to actually solve Laplace's equation using separation of variable (infinite sum of legendre polynomials) and compare that to the solution you get using method of images by expanding the latter in powers of [itex]r[/itex].

Given how easy the first method is (or should be), I suspect that the questioner expects you to do it the second way. However, the first solution should get you full credit given the ambiguous wording of the problem statement (assuming you've transcribed it word for word from your assignment sheet).
 
  • #4
thanks for the help. I got it and it was quite straightforward. I wonder why i didn't do it until now :)
thanks again.
 
  • #5


As a scientist, it is important to have a strong understanding of the fundamental principles and equations in electrostatics. In this problem, we are dealing with the concept of image charge and potential, which is a useful tool for solving problems involving conductors and charges.

To begin, let's review the concept of image charge. An image charge is a hypothetical charge that is placed at a specific location to mimic the behavior of a real charge in a given system. In this case, we have a grounded conducting sphere and a charge q placed a distance a from the center of the sphere. The image charge q' is placed at a distance a' from the center of the sphere, collinear with q and the origin. By choosing the correct distance and charge for the image charge, we can create a system that behaves as if there were two real charges present.

Now, let's look at the potential in the interior of the sphere. We know that the potential due to a point charge is given by V = (1/4πε0)(q/r), where q is the charge, r is the distance from the charge, and ε0 is the permittivity of free space. In this case, we have two charges - q and q' - and we need to consider the potential due to both of them. According to the principle of superposition, we can simply add the potentials due to each charge to get the total potential. This means that the potential at a point P inside the sphere is given by:

V = (1/4πε0)(q/r) + (1/4πε0)(q'/r')

But, we also know that the distance from the center of the sphere to the image charge q' is a' = R^2/a, since they are collinear. So, we can rewrite the potential as:

V = (1/4πε0)(q/r) + (1/4πε0)(q'R^2/a)

Now, we can simplify this equation by substituting q' = -qR/a, as given in the problem statement. This gives us:

V = (1/4πε0)(q/r) - (1/4πε0)(qR^2/a^2)

Notice that the second term in this equation is the potential due to the image charge, which we have chosen to mimic the behavior of the real charge q. So, the potential due
 

Related to Understanding Image Charge and Potential in Electrostatics

1. What is an image charge?

An image charge is an imaginary charge that is used to represent the effects of an actual charge in an electrostatic system. It is a method of simplifying the calculations in complex electrostatic problems.

2. How is an image charge created?

An image charge is created by reflecting the actual charge across a boundary, such as a conducting surface or an interface between two materials with different dielectric constants. The reflected charge is assumed to have the same magnitude but an opposite sign as the actual charge.

3. What is the purpose of using an image charge?

The purpose of using an image charge is to simplify the calculations in electrostatic problems. By introducing an image charge, the problem can be reduced to a single charge instead of dealing with multiple charges and their interactions. This can help in solving problems more efficiently and accurately.

4. What are the limitations of using an image charge?

There are a few limitations of using an image charge. One limitation is that it can only be used in electrostatic problems, and not in dynamic situations. Additionally, it can only be used in specific conditions, such as when the system has reflection symmetry and the boundary is a perfect conductor or a perfect dielectric interface.

5. Can an image charge be used in three-dimensional systems?

Yes, an image charge can be used in three-dimensional systems. However, the complexity of the calculations increases in three dimensions, and the conditions for using an image charge become more restrictive. Additionally, the accuracy of the results may decrease in three-dimensional systems compared to simpler two-dimensional systems.

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