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Gauss method for electric field evaluation. please help

  1. Sep 15, 2013 #1
    Gauss method for electric field evaluation. please help!!!!

    Hi ppl,

    I have a simple question of something that I didn't get. I guess something simple. Why in the case of Gauss solution for capacitance we add the electric field of one plate charged in positive to another charged in negative to get common electric field between two plates as you can see:

    but in the case of two charged particles we don't do it (also in the Gauss method)?

    can you please give me an examples to understand it.

    thank you very much.
  2. jcsd
  3. Sep 15, 2013 #2
    Why would you need the gauss method with 2 particles? Wat do you want to calculate with the 2 particles? The force of one on the other? The total field they produce?
  4. Sep 15, 2013 #3


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    The author of the notes used the Principle of Superposition to combine the fields from two sources: the two plates.

    You could certainly do the same for two charged particles ... it works in every situation.
  5. Sep 15, 2013 #4
    I know how to do this but if you use Gauss method the electric field of single particle will be the same as the case when you have 2 particles if you want to know the field near the one particle. you see in the case of two positive charges you will have the region near them when the electric field is zero, i.e when you put some particle there will be no force on this particle. But if you try to find this field (near one of the positive particle) with Gauss method you will not have such a case,i.e the field near this positive particle will be like there is no another positive particle. you know you enclose the surface near positive particle and find the flux... E*4pi*r^2=Q(inside)/epsilon_0.. inside this region we have only this Q of positive particle.. and find the field.. I think Gauss method should work for any case. Am I right?
  6. Sep 15, 2013 #5


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    Gauss's theorem is only useful when you have lots of symmetry. But you can find the individual fields of each of two particles (solving for all of space), and then you can add these fields together to find the combined field.

    For a detailed analysis and derivation refer to Schey's "Div, Grad, Curl and all that".

    I highly recommend it to all students of electromagnetic field theory ...
  7. Sep 15, 2013 #6
    http://www.physics.sc.edu/~crawford/phys202h/jc_chp16/lecture_5_6_gausslaw_potential.pdf [Broken]
    here you see slide7. he talks about flux near the particle that it will be positive (for positive particle) and negative for negative particle but he says nothing about superposition of fields. I didn't find nothing about that in the net, they always say that the field comes from positive to negative. Actually my question arises when I solve the problem with the charged conductive sphere with the hollow sphere around that. According Gauss you should enclose the area around the conductive sphere and moving direct to the hollow sphere. So here the situation like you have the point charge without nothing around it (like this hollow sphere around it doesn't exist).. I don't understand why is that.. It's like it's doesn't matter when you have one particle or couple of particles when you want to find field near one of the particles. it seems (according Gauss) doesn't matter if you have one particle or couple.
    Last edited by a moderator: May 6, 2017
  8. Sep 15, 2013 #7


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    In mechanics you know that the net force on a body is the vector sum of the individual forces.

    Fields obey a similar rule: if you determine the field due to each individual charge and charge distribution, the net field at each point in space is the vector sum of all of the individual fields at each location.

    This is called the Principle of Superposition; it is a fundamental property of linear vector spaces.

    See http://physicscatalyst.com/elec/charge_1.php

    Perhaps your instructor used another name for it ...
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