Aaradhana
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In the equation, Bqv=F, is q=1 Coloumb or is it equal to 1.6 x 10-19 C (like in eV)?
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Jaydude
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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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Zakee
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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.
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Aaradhana
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(Original post by Jaydude)
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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(Original post by Zakee)
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.
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Zakee
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(Original post by Aaradhana)
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.

No, charge is measured in coulombs, not in "amounts of electron". Two Coulombs of charge is different from two electrons.
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Tillybop
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(Original post by Aaradhana)
In the equation, Bqv=F, is q=1 Coloumb or is it equal to 1.6 x 10-19 C (like in eV)?
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.

(Original post by Zakee)
No, charge is measured in coulombs, not in "amounts of electron". Two Coulombs of charge is different from two electrons.
This is a good explanation of it :yy:
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Aaradhana
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(Original post by Zakee)
No, charge is measured in coulombs, not in "amounts of electron". Two Coulombs of charge is different from two electrons.
(Original post by Tilly-Elizabeth)
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 :yy:
Thanks!
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Aaradhana
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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!
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Tillybop
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(Original post by Aaradhana)
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
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Aaradhana
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(Original post by Tilly-Elizabeth)
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
Ok, I finally get that... Thanks!
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