How many electrons must be removed from an electrically neutral silver dollar to give it a charge of +2.8 µC?
Do you know the charge per electron?
number*charge per electron= total charge
solve for number.
i don't know what you mean about the charge of the electron? or what number do you times it by.
http://scienceworld.wolfram.com/physics/ElectronCharge.html
number electrons= totalcharge/chargeperelectron
q=(n*e).....(i)
where, q= charge,
n=number,
e= charge of an electron
n=q/e....(i) (from equation i)
given,q= +2.8mc= +2.8*10^-6 c
we know, charge of an electron(e)=1.6*10^-19 c
hence n= (2.8*10^-6)/(1.6*10^-19)
= 1.75*10^13
so, 1.75*10^13number of charges should ne removed.
q=(n*e).....(i)
where, q= charge,
n=number,
e= charge of an electron
n=q/e....(i) (from equation i)
given,q= +2.8mc= +2.8*10^-6 c
we know, charge of an electron(e)=1.6*10^-19 c
hence n= (2.8*10^-6)/(1.6*10^-19)
= 1.75*10^13
so, 1.75*10^13number of charges should be removed.
what? i don't get it. ISn't 1.72*10^3 the number of electrons for 2.7uC?
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1.5x10^13
To find the number of electrons that must be removed from an electrically neutral silver dollar to give it a charge of +2.8 µC, we'll need to use the formula:
Q = ne,
where Q represents the charge, n represents the number of electrons, and e represents the charge of a single electron.
First, we need to determine the charge of a single electron, which is given by e = 1.6 x 10^-19 C (coulombs).
Next, we can rearrange the formula to solve for n:
n = Q / e.
Now, let's substitute the given values into the formula:
n = (2.8 x 10^-6 C) / (1.6 x 10^-19 C).
Dividing these two values, we get:
n = (2.8 x 10^-6) / (1.6 x 10^-19).
Using scientific notation, this can be written as:
n ≈ 1.75 x 10^13.
Therefore, approximately 1.75 x 10^13 electrons must be removed from an electrically neutral silver dollar to give it a charge of +2.8 µC.