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Image charges

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data
    The configuration is that there is a charge +q a distance 3d on the z axis and a charge -2q a distance d on the z axis. The xy plane is a grounded conductor. Find the force on the charge +q.

    2. Relevant equations
    I've place image charges: -q at z=-3d and +2q at z=-d.
    the potential of the system is
    V(r) = 1/4piepsilon [q/(x^2+y^2+(z-3d)^2)^0.5 -q/(x^2+y^2+(z+3d)^2)^0.5-2q/(x^2+Y^2+(z-d)^2)^0.5 + 2q/(x^2+Y^2+(z+d)^2)^0.5]
    It think (sorry about the messy equation!

    Force is just F=qE where E is the gradient of V(r).
    The answer says that
    F=q/4PIepsilon[ -2q/(2d)^2 +2q/(4d)^2 -q/(6d)^2 ]

    How did they get that for the denominators in the brackets?


    3. The attempt at a solution
  2. jcsd
  3. Apr 15, 2009 #2
    did you apply the boundary condition that [itex]V(z=0)=0[/itex]

    also [itex]\vec{E}=-\nabla V[/itex] (u missed the negative)

    hope that helps...
  4. Apr 15, 2009 #3


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    The main thing I see here is that the force on charge +q is due to the THREE other "charges". The charge +q obviously exerts no force on itself, so you shouldn't have that in your expression.

    Find the V due to the other 3 charges and then do F=qE=q(-grad(V))
  5. Apr 15, 2009 #4
    surely by uniqueness theorem we find the potential for the entire system (i.e. all 4 charges) and then this is the same as the potential for the original system after we apply boundary conditions???
  6. Apr 15, 2009 #5


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    The potential the +q creates diverges where +q is. If you included that in your potential, locating a charge at +q is to locate the charge at a point where the potential diverges. So you really shouldn't calculate the potential for this question using all 4 charges.

    Simply put, +q can not exert a force on itself.

    The question he asked is the force exerted on +q. Not the force exerted on some other test particle.
  7. Apr 15, 2009 #6
    ok. apologies for hijacking the thread but for the simple case of a point charge q just off a conducting plate, we introduce an image charge of -q behind the plate and get V for the whole system. then to get the E for the system we can use [itex]-\nabla V[/itex]. then force is just qE - so if our original charge was at say z=a, isn't F=q E(z=a)
  8. Apr 15, 2009 #7


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    The question is "force on what"?

    That's the force on ANOTHER test charge q in the presence of that external field. (A third charge!)

    In here the question is what's the force ON +q.
  9. Apr 15, 2009 #8
    yeah i was just making up a simple example to make sure i was understanding how to do these questions properly. there's only two questions in my example though - the test charge and the image charge... im a bit confused about this third charge
  10. Apr 15, 2009 #9


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    When you say "then force is just qE"

    You mean the force on another test charge. That equation is true only for a q in an EXTERNAL field E. When I say external, I mean a field that's not created by q itself!

    Let's just make it simple.

    Say you have the SUN and the EARTH. What is the force on the Earth? It's the force due to the the gravitational field of the SUN. You don't calculate the gravitational field of BOTH and then try to find the force on the Earth. Does this make sense? The force on the Earth is due to the SUN, and the force on the sun is due to the Earth, the Earth doesn't act on itself and neither does the Sun.
  11. Apr 16, 2009 #10
    ahh ok. so for my simple example we would just get the potential of the image charge then take negative grad to get the field due to the image charge. then qE where q is th test charge and E is the field of the image charge gives the force on the test charge???
  12. Apr 16, 2009 #11


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    Yes! But you must be careful because once your test charge moves, so does your image charge! In that case the field is not "static" like it would be if there was a real (static) charge.
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