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Potential energy in an external field

  1. Nov 15, 2015 #1
    External field means field does not belong to or is not of the charges present in it,right?So,if there is a system of two point charges and these charges are present in an external field ,it means this external field is not produced by any of the charges present,right?
    Then there is a derivation for Potential energy in an external field
    consider two charges ##q_1##, ##q_2## located at ##r_1## and ##r_2## respectively in an external field E. The work done in bringing charge ##q_1##from infinity to r1 is given by ##q_1## V(r1). Similarly, the work done in bringing ##q_2## to ##r_2##, the work done is not only against the external field but also against the field due to ##q_1##.

    Work done against the external field = ##q_2## V (r2) and
    Work done against the field due to charge ##q_1##

    img64.gif
    where 'r12' is the distance between charge '##q_1##' and '##q_2##'.

    By the superposition principle for field, we add the work done on ##q_2## against the two fields.
    img65.gif

    As the path is independent of work, the potential energy of two charges q1, q2 located at r1 and r2 in an external field is given by


    potential-energy-two-charges-external-field.gif (1`)

    Here fields of ##q_1##and ##q_2## are negligible.Right?And we have to bring ##q_1##from infinity to ##r_1## first and then ##q_2## from infinity to ##r_2## we just can't pick the two charges together because by doing that we won't get the same expression for Potential energy in an external field as we got in( 1)
    Right?

    I have underlined my questions!
     
  2. jcsd
  3. Nov 15, 2015 #2

    BvU

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    Yes and Yes. The second Yes is somewhat superfluous :rolleyes:.

    Funny wording for a question in a non-homework subforum. You resent the template that much ? Answer: yes, you are correct (as far as I can judge the exercise).

    Yes, you have underlined and italicized. A bit too much of the good stuff if you ask me; legibility suffers a little. And there's that imperative "Right?" again...and yet again...

    The answers, this time, are:

    No (first "Right?") . The exercise doesn't say anything about the strength of the external field, nor about the magnitude of r12. There is nothing to ignore.

    No (second "Right?") . There are three terms in the energy and three ways you can obtain the final configuration. Physics requires that the three paths require the same energy. You sure can first position the two charges together at a cost of ##\ V = {1\over 4\pi\epsilon_0}\,{q_0 q_1\over r_{12}^2}\ ## and then move this ensemble from outside the external field to inside. But that step is just a little hard to imagine.


    Note you first mention an external field E and then an external field V. Perhaps this is an exercise you invented yourself ayway ?

    (Some external ##\vec E## fields have issues with potential at infinity, especially if they are uniform and omnipresent....)​

    --
     
  4. Nov 15, 2015 #3
    Will not the fields of these two ##q_1## and ##q_2## charges affect the :External field"?
     
  5. Nov 15, 2015 #4

    BvU

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    No they won't. That's the nice thing about fields: the contributions from various sources simply add up.
    And since we aren't being told how the external field comes about, we don't have to worry about rearranging mirror charges or anything like that. The external field is what it is and stays what it is. That's how it comes to be named 'external'. Otherwise it wouold be 'internal' to the scope under study :smile:
     
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