What is the net charge and why does it matter?

[CURIE WILLEY]

That does not matter. Draw a circle around your system and count all the charges inside. The number you get is your net charge and here it is zero.

I have replied, why I don't agree net charge is zero here. Please read it, if you find I am wrong there, pardon me and explain where I am wrong.

The position or separation does NOT matter. This is your very own definition nobody else uses. You cannot expect people to help you if you make up your own language.

If you think I have created my own definition, please specify which one I have created. And please provide correct definition for it (with reference, from where you have got that definition).

 

[Doc Al]

@DOC AL. Thank you for the reply, Sir. If we have +1 charge in one hand and -1 charge in the other hand, net charge will not be zero (as distance of separation is not zero).

False. Until you correct this statement, there's not much point of going any further. The net charge will be zero regardless of the distance.

 

[Ahsan Khan]

According to coulombs law, force between the charges is given by the equation

$F=\frac{1}{4∏ε}.\frac{q_1q_2}{r^2}$

If $q_2$ has net charge zero, it can't exert force on other charge according to me, because it has no electric field around it.

Yes if q2 is not exerting force then it can never be charged PARTICLE(remember you are saying it on the basic of Coulomb's law which is valid for point charges). But what if this is not a point,sized charged particle but a collection of charged particles(like dipole or hydrogen atom- on which you are trying to apply the definition of charge while they are not charge, for charge is exclusively the property of a(certain sub-atomic) particles you should not apply it for collection of charge))or matter with charge particles.
Do you think Coulomb would allow to use his law under such case?

 

[CURIE WILLEY]

@DOC AL. Thank you for the reply, Sir. If we have +1 charge in one hand and -1 charge in the other hand, net charge will not be zero (as distance of separation is not zero).

False. Until you correct this statement, there's not much point of going any further. The net charge will be zero regardless of the distance.

If distance of separation between the opposite charges of the dipole is zero, electric field around the system will be zero. As said always, according to me, if field is zero, net charge is zero. If distance of separation between the opposite charges is non-zero, there exists field. So, according to me, if field is non-zero, net charge is non-zero.

This has been disagreed, saying definitions of wiki is wrong. If they are wrong, it would be appreciable if correct definitions are provided (with reference, from where that definition has been extracted).

I think we are in need of knowing what actually is net charge?

So, if any particular definition from any particular reference would be appreciable. I have provided my opinion on the basis of wiki, but they are said be incorrect. So, there is a need of knowing correct definitions.

 

[ZapperZ]

Closed, pending moderation.

Zz.

 

[Dale]

This has been disagreed, saying definitions of wiki is wrong. If they are wrong, it would be appreciable if correct definitions are provided (with reference, from where that definition has been extracted).

I think we are in need of knowing what actually is net charge?

The definition of net charge was given by Doc Al in post 10.

Net charge being zero simply means that the total charge (just add 'em up) is zero: \Sigma q_i = 0

For references there are the following academic links:
http://www.physics.sjsu.edu/becker/physics51/elec_charge.htm
http://web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide02.pdf

Although it should be obvious. The term "net" simply means the sum. So your "net income" is the sum of all of your sources of income less any deductions to your income. Similarly, "net force" on an object is the sum of all forces acting on the object. Similarly, "net charge" is the sum of all the charges.

Note that $E=k\Sigma (q_i/r_i^2) \mathbf{\hat r}$ can be non-zero even if $\Sigma q_i$ is zero. Your reasoning is simply wrong, as was pointed out to you many times.