I know that when two unlike charges brought near, they will attract and cancel out each other.

So my question is what does cancel each other mean? Does they combine and form a neutral charge? I mean now they are neutral together so any of them should not be able to exert any force on a 3rd charge brought near them though they will be exerting attractive force on themselves, that is what canceling means.

But in reality they do exert force on the 3rd charge so what does this cancelling mean? It should have been like this: when two charges of opposite nature and equal magnitude are brought near will cancel each other out, now they can't exert any force on any 3rd charge.

The superposition of charges says a charge can exert force on 3rd charge even when it is combined (I mean put close together) with an equal and opposite charge.

Another argument I hear is that this cancelling means that any other 3rd charge will experience force only when placed very near, the so called neutral system, but that does not makes sense.

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    $\begingroup$ Does your textbook really say that they “cancel out”? $\endgroup$
    – Ghoster
    Apr 24, 2023 at 17:24
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    $\begingroup$ Two equal but opposite charges don't cancel each other, in classical field theory they form a dipole that will strongly radiate. In quantum mechanics they form a hydrogen like "atom", e.g. "positronium" for an electron/positron pair and they may annihilate, leading to e.g. a pair of gammas. You can, of course, try to reduce the problem to "cancelling", but that doesn't lead to actual physics. It just leads to a poor intellectual pattern that over-trivializes an actually complicated problem. $\endgroup$ Apr 24, 2023 at 17:24
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Apr 24, 2023 at 17:35
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    $\begingroup$ You need to ask the person, or maybe your teacher if that came from the textbook, or hunt down the author and ask them, in what sense the cancelling out is supposed to mean. Forming neutral atoms (or just less unbalanced charge) is a possibility. Mutual annihilation to produce light is also a possibility. But most likely they mean that the electric force fields made will cancel out and becomes less strong. In the case they are equally strongly charged, then the combination will no longer be $1/r^2$ but changes to $1/r^3$, thus essentially unobservable far away. $\endgroup$ Apr 24, 2023 at 17:40

2 Answers 2


If your charges are just abstract localized charges, then they do cancel out, because:

$$ q + (-q) = 0$$

so the total abstract charge is zero, and there is nothing there.

This is not usual, it's how we view pure dipoles, where $|q|\rightarrow \infty$ while their separation goes to zero, so the charge is indeed zero: there is NO monopole moment, but there is a dipole moment.

If you want a physical example, you can go with a positron and electron, but then you have the problem that they do more than cancel: they annihilate to 2, 3, or more photons...and there's probably a $Z^0$ channel, too.

Or go with a neutral meson, but they still have structure functions...

Or a proton and an electron, which doesn't cancel: it makes a hydrogen atom, which makes up a large portion of the (not dark) matter in the universe. Definitely not canceling out.

Sometimes it's best to let an abstraction be an abstraction.


The equal and oppositely charged particles in close proximity constitute what is called an electric dipole. It is characterized by its dipole moment, which is the defined as $\vec p = q\vec d$, where $d=\vec r_+ - \vec r_-$ is the separation vector that extends from the negative charge to the positive charge. If you place the dipole such that its mid-point is at the origin, the E-field due to the dipole at position $\vec r$ can be neatly written as $$\vec E = \frac{3\left(\vec p\cdot \hat r\right)\hat r-\vec p}{4\pi\epsilon_0r^3}.$$ This expression is valid for $r \gg d$. Note that the field magnitude is proportional to $p$ and hence $d$: the separation of the charges. As you bring the charges closer and closer, meaning $d\rightarrow 0$, the E-field (or the force on a third charge) at any point $\vec r$ gradually approaches zero. This is the sense in which the two charges "cancel out".

Note also that even for a non-zero separation, the E-field drops off with distance as $1/r^3$, faster than the $1/r^2$ variation of the field due to a single charge. This is another manifestation of the fact that the fields due to the two opposite charges partially cancel.


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