F = Bqv

Yes Q is charge of the charged particle, measured in coulombs I believe. In most questions Q is equal to 1.6x10^-19 as this will be the charge of electron or proton. ( and positron

Also don't think Q is a specific constant! It can be any number (likely very very small) which is the charge of the particle investigated.

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F = BQv.

B = flux density (this is generally constant and is any value) , Q = Charge (this could be any value). v = velocity of charged particle.

Yea, I know it's no specafic value but that was my question, is it taken in terms of electrons (as in would 1.6 x 10-19 C) be taken as one or is it like any other equation, because the value of the constant B, depends on this, right... Sorry, I think I framed the question wrong.

In the equation, Bqv=F, is q=1 Coloumb or is it equal to 1.6 x 10^-19 C (like in eV)?

No, charge is measured in coulombs, not in "amounts of electron". Two Coulombs of charge is different from two electrons.

No, charge is measured in coulombs, not in "amounts of electron". Two Coulombs of charge is different from two electrons.

If you're considering electrons then you'd use the electron charge of 1.6x10^-19C but if not then you'll consider the given charge.



This is a good explanation of it

OK, sorry to bring this up again, but what confuses me is that the electronvolt is written as eV. And then sometimes, Bqv is written as Bev (I know v is velocity here so it's different) but why is e used rather than q? I just don't get the point! It's very confusing!

When they use 'e' it means the charge of an electron, but when it says 'q' it means just charge and that they're not specifically focusing on electron charge. It's usually electrons that are considered in these situations, which is why you'll often see 'e'.

F=bev has nothing to do with eV (electron volts) because energy is not considered here, and electron volts are a unit of energy.

Hope this helps