# Coulomb interaction not affected by presence of other charges

by rajeshmarndi
Tags: affected, charges, coulomb, interaction, presence
 P: 176 One of the observation noted in connection with coulomb interaction is that, it is not affected by the presence of other charges. Why? Thanks.
 P: 234 Could you possibly tell us where you read that? It doesn't sound quite right, but perhaps I misunderstand you. From the law of superposition, we can write the effect of a set of N point charges q on a charge as $F(r) = \frac{q}{4 \pi \epsilon_0}\sum_{i=1}^N \widehat{R}_i /|R_i|^2$ Where $\widehat{R}_i$ is a unit vector in the direction of $R_{i} = r - r_{i}$. Is this perhaps what you meant? This is just due to the linearity of the electrostatic interaction - any linear system may be decomposed into a linear superposition, wikipedia has an ok writeup.
 P: 176 I read it on the twelve standard book of my state board. Also I found the same on this site. http://www.askiitians.com/iit-jee-el.../coulombs-law/ Following observations can be noted in connection with Coulomb’s interaction: (a)... (b).. (c)Coulombs interaction is not affected by the presence of other charges in the neighborhood. What exactly does it say?
 P: 234 Coulomb interaction not affected by presence of other charges I find that statement vauge too, but I would interpret it as I stated above - that the law of superposition holds.
Mentor
P: 11,872
 Quote by e.bar.goum I would interpret it as I stated above - that the law of superposition holds.
I agree.
P: 176
 Quote by e.bar.goum This is just due to the linearity of the electrostatic interaction - any linear system may be decomposed into a linear superposition
Can you explain what does linearity of the electrostatic intersection, exactly mean.

What I understand from superposition principle, is that all the charges when placed near each other behave independently of each other and just only their vector sum add up. May be this is what the statement mean.