Why doesn't charge need the equivalent of a Higgs boson?

It does, ie the photon or W/Z boson, do you mean something different?

 

To do what, exactly? Give it charge? I'm not trying to be naggy, just trying to work out exactly what you are asking : )

 

First we should look at why we need a Higgs boson for mass in the standard model.

In principle, mass need not be associated with a Higgs boson. We can write down theories in which particles have masses without needing anything like a Higgs boson. Unfortunately the standard model is not one of the theories you can write down this way. Technically speaking, the structure of the electroweak gauge symmetry means that this particular theory is inconsistent if we simply give each particle a mass. The Higgs mechanism is a way around this: the particles "start out" massless, so the symmetry is satisfied, then they obtain a mass from interacting with the Higgs field. This ends up giving a consistent theory that matches up with observations: the standard model.

You're wondering why we don't need a "Higgs boson for charge." The answer is, we can write down a perfectly consistent theory, that matches observation, in which particles simply have charge as an inherent feature. We don't need to use a trick like the Higgs mechanism to make sure we're satisfying all the required symmetries.

 

it is because EM field is infinite range field so the mediator must be massless from Yukawa style theories..... while we know that weak intereaction should be mediated by very massive boson as it is very short range.... so we require a mechenism to generate mass to them... which is Higgs mechanism....

 

The_Duck really said it all already, but I thought I might make it more explicit. Consider a massive scalar field theory (doesn't matter if you nothing about this, I will just draw a cartoon picture and make some claims you can just accept to stay with me . The Lagrangian for such a theory (from which you calculate everything else you might want to know about what happens in said theory) is

[itex]L=∂_\mu\phi ∂^\mu\phi - m^2\phi^2[/itex]

The first term describes stuff about the dynamics of the field, and the second says that the field has mass m. As The_Duck says, for fancy reasons related to the weak interactions you are not allowed to just write down something like the second term in the Standard Model. So instead you say ok, how else can we make a term that looks like that appear, without just sticking it in by hand? So you invent a new (massless) field (and set m=0 for the original \phi), giving you a fancier Lagrangian:

[itex]L=∂_\mu\phi ∂^\mu\phi + ∂_\mu\Phi ∂^\mu\Phi - y\Phi^2\phi^2 - V(\Phi)[/itex]

The second last term says the two fields interact with each other (just directly, there are no fancy boson exchanges in this theory) with some strength y. The V term says something about the potential energy of this new field, and it has to be something a bit unusual to make the field behave the way you want. That is, our goal is accomplished if the interaction term would spontaneously morph into a mass term. This would happen if [itex]\Phi[/itex] would have some constant non-zero value everywhere. You can only make it do this if the lowest energy configuration of the field is not to be zero everywhere (which is the case for the other field, [itex]\phi[/itex]), which is achieved by building the potential energy term to make this the case.

Anyway, if indeed [itex]\Phi[/itex] becomes contant everywhere ("adopts a non-zero vacuum expectation value", or VEV), say v, then the kinetic terms at the beginning vanish (since derivative of constant = zero) and the Langragian looks like

[itex]L=∂_\mu\phi ∂^\mu\phi - y v \phi^2 - V(v)[/itex]

And huzzah we have a mass term. I left out some terms which come from considering small perturbations around the VEV, which give you "higgs bosons", but this is the general idea.

So why not do something similar for charge? Well, as The_Duck says we don't need to, so why bother, but I can see that it might be interesting to check out anyway in case something cool happens. We'd probably have to think more carefully about gauge symmetries to talk seriously about this, but the following naive picture might also be useful. Consider what charge is: it tells us something about how strongly two fields interact with each other (Consider electric charge. Usually we think about two charged things, but more fundamentally it really tells about how electrically charged objects interact with photons). So our coupling constant "y" in the above theory does something similar to charge. You could replace it with yet another field, but that field would have to interact with the first with yet another coupling constant, so it looks like you would just replace one constant with another (plus a VEV), and have invented a new field, and not much would be fundamentally different in the structure of the theory. That said I haven't thought much about how this picture might be different in a proper gauge theory, but nothing interesting immediately comes to mind.

So you could do it but you'd want a good reason to go to the extra trouble. Postulating new things just costs you plausibility unless you have some specific reason to do so.

Perhaps I should add that despite the hype, I don't really think the Higgs boson tells us anything interesting at all about the origin of mass. It is not really any more illuminating to say that these mass terms comes from some fancy interaction with the Higgs field than it is to just write them straight down (although of course is important for letting us produce the mass terms at all). We are learning important things about the nature of the electroweak interactions and spontaneous symmetry breaking, but all this public stuff said about mass is beside the point as far as I see it.