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Homework Help: Need thorough explanation-electric field

  1. Jun 12, 2006 #1

    First of all, don't get scared by the size of the questions. Even if each question contains four sentances, all four of them just ask more or less the same thing. Reason is to make sure I got my point accross since I didn't know how to put question in more condensed form.

    I'm learning about electricity from the following site

    http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l4d.html [Broken]

    and all the text in quotes is from that site.

    EF ... electric field

    How can EF have net force? When talking about net forces in EF, don't we actually talk about net forces on individual objects in EF? Meaning larger the object more electric forces from that EF are exerted on it. And yes, I can see if object O1 was so large that it would cover all the nearby area ( area near source charges ) EF covers, that we could define force on O1 as net force of that EF. But if object is of very small size, then only small portion of EF affects it and net force on it could have totally different direction than net force on O1.

    At first I thought this means that net force on each excess charge on the surface of conductor is zero, but I suspect it ALSO means that there actually is no EF beneath the conductor's surface?!
    I would understand if excess charges on conductor's surface would position themself in such a way that electric forces on any of these excess charges would be zero, but it wouldn't be zero on test charges that stumbled inside that EF?!

    In case there actually isn't EF beneath the conductor's surface:

    But how can excess charges position themself in such a way that they cancel out eachother's electric forces that would otherwise create an EF beneath the surface of a conductor?

    Wouldn't it be sufficient ( for excess charge not to move ) if forces cancel eachother out at all points where there are excess charges located, but around those points electric forces could point in whatever direction? This way excess charges would still be in static equilibrium.


    Looking at the following picture

    http://img488.imageshack.us/img488/7407/u8l4c76hs.gif [Broken]

    you will notice that in straight line between the two charges there aren't any electric forces acting. I know it's true if test charge is placed in the exact middle of the two charges ( since forces from both source charges cancel eachother out ), but if test charge was nearer to one of the two charges, then there should be net force on test charge.


    If two positive electric charges are nearby, then net force at certain point of EF will be vector sum of forces F1 and F2 ( F1 will be exerted by first and F2 by second electric charges ).

    If excess charges in conductor exert forces perpendicular to surface of conductor, then whatever point we pick in EF, net force in that point will be just the result of one charge, since no two forces exerted by two different charges will be affecting same point in space?

    So if single excess charge C1 in conductor exerts force perpendicullary on an test charge T1, then this net force won't be any greater even if we put another million of excess charges around C1, since they will also exert force perpendiculary and thus none of them will affect that point in space where T1 is located?

    thank you
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jun 13, 2006 #2
    Firstly, I would advise you to learn your physics from some recognised textbook ( I recommend Resnick,Halliday ).
    Go through the chapters on Electrostatics thoroughly.
    Anyway, for the first question, I hope they mean the surface of an isolated conductor .We do not observe currents on the surface of isolated conductors, which would have been the case had there been a component of electric field parallel to the surface . Hence these surfaces are also known as equipotential surfaces.
    The second statement isn't even worded properly.
    I hope what they meant to say was that an excess charge deposited on the surface of a conductor would quickly redistribute itself at the outermost edges of the conductor.The actual proof quickly follows from Gauss' law . Have you learnt this ?
    3)Of course, there is a force actingalong the line joining the two charges.
    However they have not chosen to show it, just like they haven't shown the field lines at all points in the plane .
  4. Jun 13, 2006 #3
    I hope I don't come out sounding like a prick, but you corrected the text I quoted but in doing so you didn't explain my initial two questions.

    You did answer the third one though. :)

    And I have a textbook, but it doesn't go into any more detail than the text on that site.
  5. Jun 14, 2006 #4


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    Homework Helper

    I'll try explain the first two points. I assume you are simply not confident about your understanding from the textbook reading.

    1. The electric field is always perpendicular to a conductor's surface.
    The "net force" refers to the total force on any charge you place on the surface, whether it be real or imaginary. If you have a charged conductor, say a sphere, then there will be countless electrons on the surface. Each electron or charge will feel a force from the other charges due to the electric force (which contributes to the net force that it feels). This net force it feels is always perpendicular to the conductor's surface.

    In the perpendicular direction, the surface of the conductor can restrict or counter the electric force. That is, the conductor surface can prevent the electron from just escaping into space. However if any component of the field is parallel, it can be argued that there will be acceleration and thus charges will start moving around the sphere endlessly. This is, of course, not observed and doesn't make sense.

    2. The field inside any charged conductor under electrostatic equilibrium is always zero.
    This is proved by showing the contradiction that exists otherwise. If there is an electric field inside the conductor, then any charge within would accelerate. Accelerating charges breaks the supposed equilibrium we had originally assumed.

    If you introduce charges, then obviously these charges will move until a new equililbrium is reached. The time it takes to achieve this is usually extremely negligible. The charges on the surface will quickly rearrange themselves to preserve the equilibrium. This effect is experienced when you are trying to use your cellphone (a device reliant on electric fields) in the elevator (a conducting metal shell).
  6. Jun 14, 2006 #5
    I think ,I have replied to your questions in my earlier post , but I'll make it clearer this time.

    1)An electric field is always directed perp. to the surface of an isolated conductor. This is because the free electrons tend to align themselves in such a way, so as to alter the direction of electric field, for the sake of electrostatic equilibrium. An electric field that has a component parallel to the conducting surface can accelerate any free electrons and there is loss of equilibrium.

    The forces on a test charge within , would also be 0 and this can be proved by Gauss' law .
    Indeed, the excess charges would still be in static eqilibrium. However you seem to be forgetting about the free electrons within the conductor.They have to be in static equilibrium too. This is only possible when the excess move to the outermost surface, do you follow ?

    Oh, and you don't sound like a prick. Your doubts are intelligent and valid to the present context. :)

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