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Increase in energy in an electric field

  1. Feb 11, 2007 #1
    I was explaining the construction of Thomson's apparatus for determination of specific charge for an exam today and suddenly something popped up:

    Consider 2 oppositely charged plates held using insulated handles (They're not connected to a battery). If an electron moving with a velocity v passes perpendicular to this field then it will experience a force eE causing it to accelerate towards the positive plate. Suppose v is such that the electron escapes the volume between the plates before it can reach the positive plate. Thus the electron leaving the area between the plates would have a greater kinetic energy because the magnitude of v vector would have increased. How can one account for this energy? The plates could not have lost any energy since they are not connected to a cell or allowed to move. And the net field outside the plate system would be zero from Gauss law. Then where did this extra energy come from?!
  2. jcsd
  3. Feb 11, 2007 #2
    It's easy to see how to balance the momentum: the plates will be electrostatically pushed in the opposite direction, so claiming they aren't "allowed to move" just means the entire support is pushed as well. As for energy, which other of your statements may be faulty?

    Gauss' law is only useful given a higher degree of symmetry, since the net field doesn't tell you what the field at every point is like. When the electron travels through the finite plate system, it first approaches from a little bit closer to the negative side (so, is a little repulsed from approaching at all) and leaves from a little bit closer to the positive plate (vainly attracting it to not leave at all).

    So sure, the electron gains a second component to its motion. But since the first component of its motion is decreased in your scenario, you have not demonstrated that the electron escapes with any extra energy.
  4. Feb 11, 2007 #3


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    What happened to accounting for potential energies?
  5. Feb 11, 2007 #4
    @cesiumfrog what if the electron is approaching symmetrically from infinity. There would be a point midway between the plates where potential is zero.

    And are you sure that a considerably large apparatus would move?!

    And no the first component is not decreased. The motion in the electric field is parabolic and analogous to projectile motion under gravity so that horizontal component always remains constant. But unlike projectile motion the electron doesn't come back to 'ground' level in projectile motion. I hope I'm clear.
  6. Feb 11, 2007 #5
    Yes, but even if it approaches along that mid-surface, it will not leave along the same surface; the point is that the force as it departs will not be cancelled by the force as it approaches.

    Momentum is conserved, but for a large apparatus (compared to an electron) this produces neglegible velocity (and, in the limit, removes no energy). It is actually not relevent to your main question.

    You are only considering the motion while the electron is between the two plates (and you are approximating those plates as infinite). You should also consider the motion when the electron is not between the plates (and notice that there will still be an electric field, which also isn't completely perpendicular to the plates).

    I thought it would be redundant to quote the laws of thermodynamics, since I assumed you had realised that energy must be conserved (and therefore, the reasoning that suggested otherwise to you is flawed). Was I mistaken?
  7. Feb 12, 2007 #6
    Of course I'm aware of the conservation of energy (and hence the question!). But do the math. I've referred multiple textbooks and they all suggest that when the magnitude of the leaving velocity will be greater anyhow. But back to my original question: Where does this energy come from?

    From some of your statements I'm concluding that there must have been a net field acting on the electron against which we had to work to bring it to the initial path. This work would account for the increase in kinetic energy after moving through the field. Is that true?!
  8. Feb 12, 2007 #7
    No, I was not considering whether work must be done to put the electron on its initial path; I'm arguing that the electron's initial kinetic energy (far before the plates) will be exactly equal to the electron's final kinetic energy (far after the plates). Hence, as no energy is gained, there is also no place for it to have come from. Since this seems to contradict what your textbooks suggest, would you mind giving the references?
  9. Feb 12, 2007 #8


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    Not unless the trajectory is symmetric with respect to the mid-plane normal to the plates and the plane of motion. The electron's total energy is conserved, not its kinetic energy. After entering the region between the plates, the electron exits in a manner that puts it closer to the positive plate (than the negative one). It loses PE as a result.

    PS: Just noticed the qualifier "far", in which limit, the above statement is, indeed true.
  10. Feb 12, 2007 #9
    yup, but now I realise what the textbook might have been saying. Sounds like Mr4 just needs to find a picture of the electric field outside of a (finite) parallel plate capacitor.
  11. Feb 13, 2007 #10
    Duh, that's what i was saying in my previous posts. I'm referring some Indian textbooks namely HC verma and Nootan ISC physics (+2).

    I agree with Gokul's statement but then what does the far have to do with anything if the K.E, P.E distribution is modified.
    Last edited: Feb 13, 2007
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