A friend of mine asked me this question, that is asked in an entrance examination. It shouldn't be that difficult, but I fail to find a rigorous answer for it.

The figure shows three charges, that are fixed on a line, so that they can't move.

Electrical field

Now the question is, if $q$ is an amount of positive charge, then what are the charge of $A$, $B$ and $C$?

The possible answers are:

enter image description here

It would be nice if anyone could give a rigorous answer, because it seems at the moment that this question has to be solved by intuition, which is strange for an entrance examination.

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    $\begingroup$ Electric fields go from what kind of charge to what kind of charge? $\endgroup$
    – Michael
    Jun 1, 2013 at 12:45
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    $\begingroup$ Also: think about the total charge. $\endgroup$
    – Vibert
    Jun 1, 2013 at 12:46
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    $\begingroup$ BTW, read the homework policy if you are confused about the homework tag. It doesn't just apply to assigned homework problems. $\endgroup$
    – Michael
    Jun 1, 2013 at 12:47
  • $\begingroup$ @Michael Brown: this question can't be answered just by recalling to which direction the field lines point. You can only exclude one answer by doing that. Thanks for the homework tag tip though, I wasn't familiar with that. $\endgroup$
    – yarnamc
    Jun 1, 2013 at 19:22
  • $\begingroup$ In which institute you are studying $\endgroup$
    – Koolman
    Apr 20, 2017 at 10:11

3 Answers 3


It's not intuition.It's a problem which can be solved.

First we identify the sign of the charges. By seeing the direction of field lines we can see that the sign of charges. Field lines originate from $+ve$ and end at $-ve$ charges.

Next by Definition of Flux,

The number of field lines cutting per unit surface surface .

And Gauss' Law

The flux through a closed surface is equal to $\dfrac {Q_{enclosed}}{\epsilon_0}$

Now create imaginary spheres of same surface area enclosing each charge at a time, calculate flux and use Gauss' Law, and you'll get the ratio of charges .

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    $\begingroup$ +1 And for a test-taking situation, you can simplify it a little further. Just counting the field lines shows (via Guass's Law) that the central charge is larger than the side charges, and there's only one option satisfying that requirement. No need to wait to calculate ratios! $\endgroup$
    – Mike
    Jun 1, 2013 at 13:16
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    $\begingroup$ @Mike: This being a HW question, I prefer not to provide full solution till the answer. As stated by DavidZ $\endgroup$
    – ABC
    Jun 1, 2013 at 13:17
  • $\begingroup$ The problem is nicely stated as it does not require to remember the direction of field lines between charges. $\endgroup$
    – babou
    Jun 1, 2013 at 15:50
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    $\begingroup$ Well, the problem does not display all field lines. There can be lines going from A to C without passing through B. Also, there can be lines going from A to infinity. As a consequence, you cannot really count lines and use Gauss' law. $\endgroup$
    – fffred
    Jun 1, 2013 at 16:29
  • $\begingroup$ Thanks, I was not thinking about Gauss law, because this question can normally be answered with knowledge of secondary school, because it comes from an entrance examination and Gauss law isn't taught in secondary school here. $\endgroup$
    – yarnamc
    Jun 1, 2013 at 19:19

Field lines originate from +ve and end at -ve charge. So its clear A and C are -ve and B is +ve. OK?.... now no. of lines originating from B over total solid angle 4pi is 10 no. of lines ending at A and C over total solid angle 4pi is 5 so |B|/|A|=10/5=2, and |B|/|C|=10/5=2 so B is 2 times A and C and +ve>>>2q A nd C are -q so option (d) is correct


Ans . Is D ,because B is + Ve (lines are coming from it)and A,C are negative( liner are merging in them) and no. Of lines
OnA,C are 5 in no. And onB is 10 double in magnitude ,so if A,C charge is q then charge on B is to be 2q. Hence


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