# Electrostatics: what is the intensity of a charge at the position of the charge?

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.

• 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). – Rudyard Nov 8 '16 at 14:17
• 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? – safeer khan Nov 8 '16 at 14:34
• 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 $k\frac{Q}{r^2}=-E/3$ – Rudyard Nov 8 '16 at 14:47

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