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I'm solving a problem: Two charges are given such that the distance between them is $r$ and intensity at point of other is $E$. If $q$ and $-3q$ are the charges, then the intensity at point of $-3q$ will be?

Please give some hints as I'm completely confused. I thought as there are two charges then the field of both the charges will act on unit positive test charge so the total intensity will be

$E=K(E_1+E_2)$

where $E_1$ and $E_2$ are intensities due to $q$ and $-3q$ respectively and $K$ is the electrostatic constant. But distance between $q$ and test charge is $0$ and so $E_1$ becomes infinity.

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    $\begingroup$ So you have two charges: q and -3q, at a distance r. Are you sure the question isn't asking for the force on the charge -3q caused by charge q? (It doesn't make sense to talk about the electric field induced by a point charge on top of the point charge itself; as you mentioned, the electric field approaches infinity as yo get closer and closer). $\endgroup$ – Rudyard Nov 8 '16 at 14:17
  • $\begingroup$ Actual question:two point charge Q and -3Q are placed some distance apart. If the electric field at the location of Q is E then at the locality of -3Q it is? $\endgroup$ – safeer khan Nov 8 '16 at 14:34
  • $\begingroup$ I believe the answer you're probably looking for is $-E/3$, because the electric field is directly proportional to source charge. Call the distance $r$. At Q the electric field caused by -3Q is $E=k\frac{-3Q}{r^2}$ (where $k$ is some constant not important here). At -3Q the electric field cause by Q is $k\frac{Q}{r^2}=-E/3$ $\endgroup$ – Rudyard Nov 8 '16 at 14:47
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A (semi-)formal definition for the electric field, $\vec{E}$ at a point is $$ \lim_{q'\rightarrow 0}\frac{1}{q'}\vec{F}_{\mathrm{on\ q'}}.$$

A point charge does not contribute to the $\vec{E}$-field at its own location, but only other charges do. In your situation, with two point charges, you can find the force on a charge and simply divide by that charge. That will give you the $\vec{E}$ at that location.

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The forces on the 2 charges are equal and opposite (Newton's 3rd Law). The electric field strength is $E=F/Q$. Therefore $\frac{E_1}{E_2}=\frac{Q_2}{Q_1}$.

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