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It is well known fundamental behaviour that, oppositely charged bodies attract each other (I don't know whether it applies also for charges of equal magnitude or not), and identical charges repel each other.

It is also well known that, a system of oppositely charged bodies with equal charge in magnitude, has zero net charge.

If a system has oppositely charged bodies, with equal charge in magnitude, it would imply that, there will be zero net charge, and the vector field's had got cancelled each other, like around a current carrying conductor, there is no electric field because, charge on current carrying conductor is zero, as one electron enters the conductor, the other will be leaving the conductor. Now, if there is no field, how could the oppositely charged bodies (of equal charge in magnitude) attract each other, they should not feel any force, isn't?.

Taking into account all the above statements, would it imply that, two oppositely charged bodies (of equal charge in magnitude) attract each other?

[All statements made, are up to my view. Any correction advisory is welcome]

LINKS

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    $\begingroup$ I am sorry but I will also vote down, because the way you frame your question shows that you know nothing about vector fields or their addition. One cannot substitute a 101 course in physics with an answer to a question. $\endgroup$ – anna v Nov 3 '13 at 14:16
  • $\begingroup$ @annav.Thank you for giving reason. I would correct myself if you can suggest any corrections in the question, or else if you can edit the question. $\endgroup$ – Immortal Player Nov 3 '13 at 14:24
  • $\begingroup$ see this answer by josh physics.stackexchange.com/questions/57431/… . The electric fields of positive and negative charges add vectorially into a larger force because the vector of E points in the same direction from positive to negative . They cannot cancel. See also this hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c2 $\endgroup$ – anna v Nov 3 '13 at 17:44
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Taking into account all the above statements, would it imply that, two oppositely charged bodies (of equal charge in magnitude) attract each other?

Yes. Two opposite charges will always attract, independently of their magnitudes. That's what Coulomb's law says: $F=\dfrac{kq_1q_2}{r^2}\implies F=-\dfrac{k|q_1q_2|}{r^2}$. If they are equal in magnitude: $F=-\dfrac{kq^2}{r^2}$

If a system has oppositely charged bodies, with equal charge in magnitude, it would imply that, there will be zero net charge, and the vector field's had got cancelled each other

No. Total net charge doesn't mean zero field, because there is some separation $d$ between the charges.

The field for an individual charge is: $E=\dfrac{kq}{r^2}$.

And the superposition principle states that fields add up: $E_{\mathbf{total}}=E_1+E_2$.

This means that, if we have two opposite charges of equal magnitude:

$$E_{\mathbf{total}}=E_1+E_2=\frac{kq}{r_1^2} - \frac{kq}{r_2^2}$$

Unless $r_1=r_2$ is zero (that is, $d=0$), $E$ will never be zero. Intuitively, you can think that each field tries to cancel, but they can't because there is some displacement. The smaller the distance between the charges, the cancellation will be better. If the charges are largely separated (one charge here and other in the Moon), you would have two independent fields.

See for example the field of a dipole:

enter image description here

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  • $\begingroup$ @jinawee.The question is about electric field or force acting on either charge, so you need to consider $d$ as $r$ in the electric field equation. Here, in your answer, you are finding force on some other test charge placed at a distance $r_1$ from one charge, and $r_2$ from other charge. So, the above calculation would be wrong. $\endgroup$ – Immortal Player Nov 3 '13 at 14:53
  • $\begingroup$ @CURIE First of all, I calculated the force that each charge would feel is: $F=\frac{kq^2}{d^2}$, in this case it clearer to use d instead of r, because in the second part r has a different meaning (notice that the force on some other test charge Q would be $F=\frac{kqQ}{r_1^2} - \frac{kqQ}{r_2^2}$). So there's nothing wrong. $\endgroup$ – jinawee Nov 3 '13 at 15:09
  • $\begingroup$ ."....notice that the force on some other test charge Q....." What I thought is, we need no test charge.I really appreciate your way of solving this problem.I have no arguement with it. I hope you would think like this, we need to find force or field on either charge.You are right with the expression of force on each charge, but if $d$ is zero, force on each charge will be infinity, I don't know whether we get value using limits or not. Then, your arguement of field being zero, when $d$ is zero falls here. Because, when such a infinite force is acting on the charges, there must be...... $\endgroup$ – Immortal Player Nov 3 '13 at 17:23
  • $\begingroup$ @jinawee....net field associated, when $d$ is zero.So, force being infinite, and field being zero, when $d$ is zero are contradicting from your answer, indicating that, method used by you is not correct. $\endgroup$ – Immortal Player Nov 3 '13 at 17:40
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    $\begingroup$ @CURIE I'm starting to understand what you mean. First of all, it makes no sense to have to charges at distance d=0, you would need infinite energy to get them so close. Another problem is that you CAN'T use $F=\frac{kq^2}{r_1^2} - \frac{kq^2}{r_2^2}$ for one of the charges, because that means that the particle is exerting a force to itself (nonsense). So the field decrease far off the charges but it increases between them (I migth add a plot tomorrow to see this). $\endgroup$ – jinawee Nov 4 '13 at 0:07

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