As we have developed the idea of electric field lines from Coulomb's law. If two equal and opposite charges are in plane, all the effect of electric field vectors adds to line joining the two points, which means all the electric field lines of forces contribute to Coulomb's force.

When more than two charges are in plane, the electric field lines distribute between charges, which means all electric field lines do not contribute to apply a force between the charges. It should be reduced by this analogy. But we use Coulomb's force the same, whether two charges or more charges. Why? It should be reduced if more than two charges are in plane by distribution of electric field lines.


  • How can charge $q_1$ apply the same amount of force, be it one other charge in space or multiple ($n$ charges)?
  • How is it possible?
  • $\begingroup$ In some cases the force is reduced. Remember that the forces add vectorially so that if you have two equal (in magnitude) and opposite forces, they result in a net force of zero. If that doesn't answer your question please try to clarify what you mean so that I can hopefully give you an answer. $\endgroup$
    – user273751
    Commented Oct 23, 2020 at 7:01
  • $\begingroup$ @PiKindOfGuy Electric Field Lines/Intensity (E) span complete space radially outward by charge q1. If i placed opposite q2 charge in space at some distance all E of q1 curved and attract q2 not just straight line. Because when i added upward and downward E vector there resultant is also straight line. I can come up with this result all E contribute to attract q2. not just E of straight line. $\endgroup$
    – 123
    Commented Oct 23, 2020 at 7:11
  • $\begingroup$ Because the above and below forces also add up to straight line. so they are also contributing in attraction. What if i placed q3 charge. Now q1 electric field not all go to q2. because in their path there is another charge. Coulomb force should distribute between charges. $\endgroup$
    – 123
    Commented Oct 23, 2020 at 7:14
  • $\begingroup$ So you're claiming that q3 would "steal" at least part of the electric field lines going from q1 to q2 and therefore q1 shouldn't attract q2 as if q3 weren't there? $\endgroup$
    – user273751
    Commented Oct 23, 2020 at 7:18
  • $\begingroup$ @PiKindOfGuy Yes that's what i meant. $\endgroup$
    – 123
    Commented Oct 23, 2020 at 7:30

3 Answers 3


If I'm reading your question and your comments correctly, you seem to be thinking that if there are two charges $q_1$ and $q_2$, then the force on $q_2$ due to $q_1$ is because of all the field lines coming out of $q_1$. That's not true.

To find the force on $q_2$ due to $q_1$ from the field lines, you first need to find the electric field from the field lines. The direction of the electric field is just tangent to the field line at the point where $q_2$ is located. To find the magnitude of the electric field at the point where $q_2$ is located, you need to follow a procedure. If you want to know why this procedure works, I suggest you read up on Gauss's law. If you have a convention of $n$ lines per coulomb where $n$ is a very big number, then take a very small area element $dA$ centered about the point and count the number of field lines passing through $dA$. The electric field at that point is $$E = \frac{\text{# of lines passing through } dA}{n \epsilon_0 dA}$$ Then the force on $q_2$ is just $q_2E$. The larger you take $n$ to be, the smaller you are allowed to take $dA$ to be. So really you can take $dA$ to be as small as you like. The takeaway from this is that the electric field strength and hence the force on $q_2$ due to $q_1$ is only dependent on the field lines of $q_1$ very close to $q_2$. All the field lines of $q_1$ are not responsible for the force on $q_2$.

  • $\begingroup$ Hi, What about the analogy of Vercassivalunos comment. $\endgroup$
    – 123
    Commented Oct 25, 2020 at 18:51
  • $\begingroup$ Which analogy? I don't see him making an analogy. $\endgroup$ Commented Oct 25, 2020 at 19:26
  • $\begingroup$ That's why i asked. Because it didn't find any scientific description in this. That's why i asked do you agree with his analogy. $\endgroup$
    – 123
    Commented Oct 25, 2020 at 20:11
  • $\begingroup$ I know Electric Field Lines (E) are just representation of vector field which just give us analogy of direction and magnitude of E (analogous to acceleration in mechanics) depend on source charge, responsible of acceleration of other charges if it is free to move. But this E idea gives us a way of calculation a physical phenomenon. I asked question i wonder how q1 act same force on n charges in space. In case of gravity earth apply different force on different mass to keep $\vec{g} = 9.8 \frac{m}{s^2}$ .It is mystrey. $\endgroup$
    – 123
    Commented Oct 25, 2020 at 20:21
  • $\begingroup$ $q_1$ doesn't apply the same force. Force on $q_2$ and $q_3$ due to $q_1$ would be $q_2E$ and $q_3E$ respectively where $E$ is the electric field due to $q_1$. The force will be different for different test charges but the $E$ due to $q_1$ is the same. Much like how force of Earth is different on different masses but the $g$ due to Earth is the same. $\endgroup$ Commented Oct 25, 2020 at 20:46

Firstly, the electric field lines just give us just the direction of electric force. When a third charge is introduced, As you mentioned the field lines bend and the third charge may steal some of the original field lines. But what we forget is that the magnitude of $\vec E$ at any point would also have changed.

Hope that clears why stealing of field lines need not reduce the force

Now, we find the resultant Force on any body due to a set of forces by vector addition of forces. In the three charges case, (let's call them $q_1$ , $q_2$ , $q_3$). The force on any charge (say $q_1$) would be the vector sum of forces from $q_3$ and $q_2$. This is just a property of vectors and forces in general. Nothing special for Coulomb's law

fig 1
fig (1)

fig 2

in both of these figures, a charge $q_3$ is introduced near a pair of charges $q_1$ and $q_2$. In both these cases, the field lines will be stolen but The force on $q_1$ decreases in fig(1) but increase in fig (2).

Another thing I should've mentioned is that when we say we can do vector addition to find the force, we are assuming that the charges won't move from their position due to these forces. $q_1$ will remain where it is and so does $q_2$ and $q_3$.


The field lines are not a real physical quantity. Your Idea of field lines seem to be as if something is coming along those lines and hitting the charges and transferring force. That is not the case. Field lines are just lines showing the direction of force. Direction only.

To make you understand that you cannot count the field lines falling on a charge to find the force, Let us consider 2 charges. enter image description here

enter image description here

In first figure I've drawn only 6 field lines. So accordingly, the force would be due to 6 lines. In the next image there are more lines drawn. So according to your Idea the force should increase. This is not at all true. How can the force depend on number of imaginary lines we draw.
What we should do is measure the Field at a point and use it to find the force.

enter image description here

This diagram shows the force at any point due to both charges. see how if we draw a curve along their tangents we get the Field lines. Thats all there is in a field line. It does not give the magnitude of force and at any point, there would be only one direction for these lines. You need not add all the lines falling on it.

  • $\begingroup$ Hello @Rishab Navaneet . Pls explain briefly why magnitude of E does not change when another charge in place. $\endgroup$
    – 123
    Commented Oct 24, 2020 at 9:41
  • $\begingroup$ Let say if q1 has 10 electric field lines (I know this is just for vector addition purpose but it represent E magnitude and direction in space) and q2 is placed. All 10 E lines of q1 responsible for applying 10N force on q2 not just line joining the two charges. If you add above vector and same below vector the is once again same straight line. With this fact i can come up with result all the E lines create 10N force force q1.If another charge q3 is placed now E lines distribute, it should not be available all 10 E for both charges.By this analogy q1 10N should be distribute between q2 and q3. $\endgroup$
    – 123
    Commented Oct 24, 2020 at 9:53
  • $\begingroup$ I told you not to find the field from field lines. Use the field to find field lines. Your Idea of field lines is like that of many balls following those lines. All of them comes and hits the charge and all those forces add up. That is not correct. Field lines are just lines drawn in space that show the direction of Electric field at that point. All the field lines meeting at one charge doesn't mean all those lines exert separate separate forces. I'll edit my answer. $\endgroup$ Commented Oct 24, 2020 at 13:47
  • $\begingroup$ In my analogy number of lines can only be increase when magnitude of charge increase. If E is only for direction why we find magnitude of force by test charge every point in space due to two charges? $\endgroup$
    – 123
    Commented Oct 24, 2020 at 16:20
  • 1
    $\begingroup$ @RishabNavaneet Not necessarily, if you choose a convention of picking $1/ \epsilon_0$ lines for each coulomb of charge, the areal density of field lines gives the strength of the field. If you choose any other convention of n lines per coulomb (n is sufficiently big), then the density is proportional to the field strength upto a proportionality constant, so it still gives the field strength. $\endgroup$ Commented Oct 25, 2020 at 17:11

The electric field of a given charge configuration tells you what force it exerts on an additional test charge placed in the configuration, not how the charges in the configuration interact with each other (you can also get that information out of the charge and field line configuration, but it's not what the field lines are supposed to tell you, so that's slightly more involved).

So if you have three charges $q_1,q_2,q_3$, then to find the force exerted by $q_1$ on the other two charges, do not consider the field lines of the entire charge configuration. Instead, take only the field lines of $q_1$ alone, and ignore how $q_2,q_3$ alter the field lines. These unaltered field lines going out radially from $q_1$ tell you the force exerted by $q_1$. Look at the line density at the position of $q_2/q_3$. The electric force exerted on the two charges is proportional to this density and is parallel to the lines. And since we're talking about the density of the field lines of $q_1$ alone, without the alterations caused by the other two charges, this density obviously doesn't depend on the presence of other charges.

Of course, $q_2$ and $q_3$ also exert forces on each other. To find those forces, you have to do the same thing: consider only the field lines of an isolated charge, and use those to find the forces it exerts. Or to get the total force exerted on $q_3$, you can take the combined field lines of only $q_1$ and $q_2$, and then find their density at the position of $q_3$. But whatever force you want to find, always use field configurations which don't include the field lines of the charge you want to examine. Because the field lines of a charge are supposed to be used to find forces exerted by said charge, not on it.

  • $\begingroup$ Hi @Vercassivelaunos . Thanks, is there any book at which i can find the detail what you shared about configuration. $\endgroup$
    – 123
    Commented Oct 25, 2020 at 10:58
  • $\begingroup$ I don't know any, since I learned from lectures, not books. But any good introduction of electric fields will mention that the field $\vec E$ is by definition the quantity that characterizes the electric force experienced by an external charge $q$ placed into said field via $\vec F=q\vec E$. The field lines we usually draw are just a graphic representation of this quantity. $\endgroup$ Commented Oct 25, 2020 at 12:42

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