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Electric field potential

  1. Nov 14, 2009 #1
    Electrons are emitted from cathode C and accelerated towards anode A.

    http://img526.imageshack.us/img526/7655/efield.jpg [Broken]

    Anode is earthed. Electric field strength along CA is shown.

    Question:
    Arrow of direction of electric field along CA.

    My ans would be left to right as I thought e-fields are always from high to low potentials?

    But the answer says opposite as electrons experience an electric force opposite in the direction to the electric field (which I know is correct).

    Anyone can help answer my querry? Why isnt looking at high to low potential in the graph correct?

    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 14, 2009 #2

    Delphi51

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    The cathode is negatively charged with extra electrons.
    The anode is positively charged with a shortage of electrons (extra protons).
    The electric field comes out of a positive charge and goes into a negative charge so it goes from the positive plate toward the negative plate. This is by definition of the electric field direction.
    The E field is approximately constant between the plates; it doesn't fade away like the graph shows.
     
  4. Nov 14, 2009 #3
    Ok sorry. I should clarify myself. They are not two parallel plates but rather the cathode is like a point. Sorry if I drew it wrongly with paint. Anyway, I redrew it.

    I thought e-fields are always from high to low potentials?
     
  5. Nov 14, 2009 #4

    Delphi51

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    The field will still be constant along the line through the centers of the electrodes.
    Yes, the field lines go from the high potential to the low. Positive is taken to be high, negative low. This is a definition. The pioneers could have chosen the field direction to be the other way round.

    Note, too, that there is no absolute potential. Rather, we have a potential difference between the two electrodes because they are connected to a battery or power supply. This moves some electrons from anode to cathode so there are net charges, and that causes an electric field from the positive charges to the negative ones.
     
  6. Nov 14, 2009 #5
    Yes, but isnt there a contradiction? The potential at C is higher than A, yet the field lines are from A to C?

    Sorry if Im quite confused on this.
     
  7. Nov 14, 2009 #6

    Delphi51

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    No, the potential at A is positive compared to C, so A has a higher potential.
    Say the Cathode is grounded and A has a potential of 1000 Volts. Then A is 1000 V higher than C. The E field between is E = V/d = 1000 divided by the distance.
     
  8. Nov 14, 2009 #7
    But the graph says C is at a higher potential. The graph shows potential varying from C to A for 4 cm. Furthermore, the anode is grounded.
     
  9. Nov 14, 2009 #8

    Delphi51

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    Ah, was the graph given in the question? I just assumed the diagram was part of your answer and that it is wrong. The graph seems wrong to me because the E field along the line between two electrodes should be a horizontal line - constant E.

    The graph does not say the potential is higher at C. It shows the electric field, not the potential.

    If the graph is supposed to show V vs distance rather than the labeled E vs distance, then it is incorrect because it shows the cathode positive. No electrons would be emitted from a positive electrode. It would make sense if the vertical scale is negative upward rather than the standard positive upward.
     
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