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Homework Help: Method of images, charge q inside conducting sphere

  1. Oct 26, 2017 #1
    1. The problem statement, all variables and given/known data
    Using the method of images discuss the problem of a point charge q inside a hollow grounded conducting sphere of inner radius a.Find
    (a) the potential inside the sphere
    (b) induced surface-charge density
    (c) the magnitude and the direction of force acting on q
    is there any change of the solution i f the sphere is kept at a fixed potential V? If the sphere has a total charge Q on it?

    2. Relevant equations


    3. The attempt at a solution
    parts a) and b) were easily solved
    By using method if images:
    ##q'=\frac {-qa}{y}##and ##y'=\frac{a^2}{y}##
    the potential at point p is ##Φ(p)=\frac{q}{4πε|x\vec n-y\vec n'|} +\frac{q'}{4πε|x\vec n-y'\vec n'|}##
    the magnitudes of q and q' are chosen so that the potential at the surface of the sphere becomes zero.
    the result will be by using spherical coordintes:##Φ(x)=\frac {q}{4πε} [ \frac {1}{\sqrt {y^2+x^2-2yxcos(\gamma)} } -\frac{1}{\sqrt {a^2+y^2x^2/a^2 -2yxcos(\gamma)} }]##



    Now the surface charge density is## σ=εE=-ε\frac {\partial Φ(x)}{\partial x}|_{x=a} ##that will leads finally to
    ##σ=\frac {qa/4π (y^2/a^2 -1)}{(y^2+a^2-2yacos(\gamma))^{3/2}} ##


    I really stopped at part (c)
    the easiet way is to find the force between q and q' by using Coulombs equation.
    the answer will be ##|F|=(q^2/a^2)(a/y)^3(1-a^2/y^2)^{-2}##
    but when I tried solving it by calculating the total force acting on the surface of the sphere I got a different answer!!
    the force acing on an element area of the sphere is ##F=(σ^2/ε)da## where ##da=asin\gamma d\gamma dφ##
    Then ##F=(1/ε)\int_0^π\int_0^{2π}asin\gamma d\gamma dφσ^2=##
    ##\frac {q^2a^3}{16π^2ε} \int_0^π \frac { (y^2/a^2 -1)^2sin(\gamma) d\gamma}{(y^2 +a^2 -2yacos(\gamma))^3} \int_0^{2π}dφ=##
    ##\frac {(q^2a^3)}{8πε}\int_0^π \frac { (y^2/a^2 -1)^2 sin\gamma d\gamma}{(y^2 +a^2 -2yacos\gamma)^3}##
    Let ##A=(y^2 +a^2-2yacos\gamma)##
    ##dA=2yasin\gamma d\gamma##
    then##F=(q^2a^3/8πε)(y^2/a^2-1)^2(1/2ya)\int_{y^2+a^2-2ya}^{y^2+a^2+2ya}dA/A^3##
    The answer finally will be ##\frac {q^2a^2}{-32πε}(1/a^2 -1/y^2)[1/(y+a)^4 -1/(y-a)^4]##
    which is different from the answer above, what went wrong?
     

    Attached Files:

  2. jcsd
  3. Oct 26, 2017 #2

    kuruman

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    Why did you pick the derivative with respect to x? Why not y or z? Remember that σ is a scalar and E is a vector. Check your textbook to see what the actual expression is.
     
  4. Oct 26, 2017 #3

    TSny

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    I think the OP is correct here. The notation seems to be that of Jackson's text. x represents the magnitude of the position vector, ##\mathbf x##, of the field point, not the ##x## coordinate of the point. Most people would use r where Jackson uses x. Definitely can be confusing.
     
  5. Oct 26, 2017 #4

    kuruman

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    OK, I think I see it now. It is definitely confusing in view of OP's posted equation
    ##Φ(x)=\frac {q}{4πε} [ \frac {1}{\sqrt {y^2+x^2-2yxcos(\gamma)} } -\frac{1}{\sqrt {a^2+y^2x^2/a^2 -2yxcos(\gamma)} }]##
    in which the meaning of ##x## on the left is different from the meaning of ##x## on the right. Furthermore, despite OP's assertion, this equation is not in spherical coordinates. If ##x## on the left is to be interpreted as ##r##, then on the right side ##x## appears in the wrong places and ##y## has no place at all.
     
  6. Oct 26, 2017 #5

    TSny

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    The ##x## on the left should be in bold face as it represents the position vector of the field point. The ##x## on the right is the magnitude of this vector.

    ##\mathbf y## is the position vector of charge ##q## and ##y## is the magnitude of this vector. ##\gamma## is the angle between the two vectors ##\mathbf x## and ##\mathbf y##.
     
  7. Oct 26, 2017 #6

    kuruman

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    Thanks. I see it now.
     
  8. Oct 27, 2017 #7
    Sorry I didn't specify the vectors and the magnitudes.They are like how @TSny mentioned them.
    I meant by E here the magnitude of E and not the vector. Sorry I was not clear.
    So what goes wrong in part (c) what I did wrong?
     
  9. Oct 27, 2017 #8

    TSny

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    You need to take into account that the force on a patch of area ##da## is a vector quantity. The direction of the force is different for different patches.

    The magnitude of the force on area ##da## is ##\frac{\sigma^2}{2 \varepsilon} da## with a factor of 2 in the denominator.

    [Edit: Also, check your expression for the area element ##da##. It doesn't have the dimensions of area.]
     
    Last edited: Oct 27, 2017
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