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Homework Help: Method of images

  1. Jul 22, 2009 #1
    1. Point charge in the presence of a grounded conducting sphere given the boundary conditions phi(x)=0 at r=a (V=0) and at r=infinity (V=0)

    2. The equation I used phi(x)= (q/abs(x-y))+(q'/abs(x-y))=(q/abs(x*nhatsubx-y'nhatsuby))+ (q'/abs(x'*nhatsubx-y'*nhatsuby')

    3. The equation listed above is what I came up with at r=a q/abs(anhat-yn'hat)+q'/abs(a*nhat-y'*n'hat)=0, my next equation was q/a*abs(nhatsubx-(y/a)nhatsuby')+q'/y'abs(n'hatsuby-(a/y')nhatsubx). After solving they come up with the ratio (q/a)=-(q'/y') and (y/a)=(a/y'). Is it because of the symmetry that they are using these ratios as opposed to going through the rigormorale of law of cosines?
    1. The problem statement, all variables and given/known data
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jul 22, 2009 #2


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    Hi phenolic, welcome to PF!:smile:

    This forum supports LaTeX, which helps in writing clearer equations. For example, I assume by the above equation you mean:


    where [itex]\textbf{x}[/itex] is the vector from the origin to the field point, [itex]\textbf{y}[/itex] is the vector from the origin to the point charge [itex]q[/itex], and [itex]\textbf{y}'[/itex] is the vector from the origin to the image charge [itex]q'[/itex]? Is this what you mean?

    Notice how I explained what the variables in my version of the equation meant, and how much clearer the meaning of the equation became?:wink:

    Do you mean


    where [itex]\textbf{n}[/itex] is a unit vector in the direction of [itex]\textbf{x}[/itex] and [itex]\textbf{n}'[/itex] is a unit vector in the direction of [itex]\textbf{y}[/itex]?

    Are the "they" you are referring to, Jackson's Classical Electrodynamics?

    If so, it's not really symmetry that they are appealing to, it's simply looking at the above equation and recognizing that the choices [itex]\frac{q}{a}=-\frac{q'}{y'}[/itex] and [itex]\frac{y}{a}=\frac{a}{y'}[/itex] automatically will make [itex]\Phi(x=a)=0[/itex]. The uniqueness theorem guarantees that this solution is the only possible solution, and so there is no need to rigorously go through the math using the law of cosines.

    They do however appeal to symmetry in the previous step when they assume that [itex]\textbf{y}'[/itex] points int the same direction as [itex]\textbf{y}[/itex].
  4. Jul 22, 2009 #3
    Thanks that makes sense and I will look into LaTeX since I am not familiar with that, but to your questions yes and that does make sense. I guess I overlooked the uniqueness theorem but the uniqueness theorem does not apply to all problems like this does it?
  5. Jul 23, 2009 #4
    Any time your using the method of images to solve for a potential, a uniqueness theorem is exactly what lets you solve the problem. Since the uniqueness theorem states that there can only be one potential that satisfies both Poisson's Equation and the given boundary conditions in the specified region, then if you can find one that fits the bill by clever guess (which is what we're doing with the image charges that we are placing outside of the region of interest) it hast to be THE solution.

    If you solve the problem by some other method (separation of variables, say) then the uniqueness theorem doesn't play such a prominent roll, but it still tells you that the one solution you find is indeed the only solution.
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