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Homework Help: In which direction is the electric field of the point charge

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data

    There is a negative point charge between 2 oppositely charged plates. Positive at top and negative at bottom. The electric field runs downwards from pos to neg.

    In which direction is the electric field of the point charge.

    2. Relevant equations



    3. The attempt at a solution

    I have calculated the value of the electric field and have a positive result. Does this mean that the electric field of the point charge goes upwards because the negative value of the point charge is attracted towards the positive plate.

    Any help with field directions would be greatly appreciated.
     
    Last edited: Feb 1, 2010
  2. jcsd
  3. Feb 1, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi spiderlegs! Welcome to PF! :smile:

    Can you check the wording of the question?
    Do you mean the force on the point charge?

    Do you mean the electric field (fromn that plates) at the point charge?

    (the electric field of the point charge is simply radial, pointing outwards from the charge :wink:)
    The test charge for measuring fields is always positive … you ask what happens to a positive unit charge.

    So a positive charge always is attracted along the arrows, and a negative charge is attracted the other way. :smile:
     
  4. Feb 1, 2010 #3
    Hello there, thanks for your speedy reply. Sorry for being so vague, it wasn't deliberate. However, the question is rather much more complex.

    There is a negatively charged oil drop value -6.4 x 10^19 placed 2mm horizontally left of a fixed charge Q, value 2 x 10^11. They are both midway between oppositely charged plates. Positive plate at top 500V.

    Calculate the horizontal and vertical components of the electric field at the oil drop due to the plates and the fixed charge Q. Indicate the approximate direction of the net electric field on a diagram.

    Attempt at a solution:

    The field at the oil drop due to Q has components ε_y=0 and ε_x=-Q/(4πℇ_o r²) .

    ε_x= -45000 N C¯¹

    The field at the oil drop due to the plates has components ε_x=0 and ε_y=-(∆V/∆y)

    ε_y=62500 N C¯¹

    For the diagram I have drawn the vertical component in the positive y direction and the horizontal component in the negative x direction with the net force between the two pointing NW.

    Does that help?
     
    Last edited: Feb 1, 2010
  5. Feb 1, 2010 #4

    tiny-tim

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    Hello spiderlegs! :smile:
    (btw, why are you using ε? E is usual for an electric field … and ε is terribly difficult to write in an exam, if you want to make it look different from ε0 :wink: also, try using the X2 tag just above the Reply box)

    ah! :smile:

    Yes, your solution looks fine.

    The field at the oil drop due to the plates is down, and the field due to the positive Q is away from Q … total down and away.

    The negative oil drop will therefore go up and towards.
     
  6. Feb 2, 2010 #5
    Thanks very much for your help ..
     
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