Potential energy in an external field

[gracy]

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 $q_1$

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.

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

(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!

 

[BvU]

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?

Yes and Yes. The second Yes is somewhat superfluous .

consider two charges q1 q_1 , q2 q_2 located at r 1 r_1 and r 2 r_2 respectively in an external field E. The work done in bringing charge q1 q_1 from infinity to r1 is given by q1 q_1 V(r1). Similarly, the work done in bringing q2 q_2 to r 2 r_2 , the work done is not only against the external field but also against the field due to q1

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).

Here fields of q1 q_1 and q2 q_2 are negligible.Right?And we have to bring q1 q_1 from infinity to r 1 r_1 first and then q2 q_2 from infinity to r 2 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!

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...)​

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[gracy]

The exercise doesn't say anything about the strength of the external field, nor about the magnitude of r12

Will not the fields of these two $q_1$ and $q_2$ charges affect the :External field"?

 

[BvU]

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