How many electrons must be removed from an electrically neutral silver dollar to give it a charge of +2.1 µC?
Each electron removes e = 1.6*10^-19 C
You figure it out.
N*e = Q
Solve for N
To determine the number of electrons that must be removed from an electrically neutral silver dollar to give it a charge of +2.1 µC, we need to use the formula that relates charge to the number of electrons.
The elementary charge is the charge of a single electron and is given by:
e = 1.6 x 10^(-19) C (Coulombs)
Given that the charge of the silver dollar is +2.1 µC, we need to convert it into Coulombs before proceeding further.
1 µC = 1 x 10^(-6) C (Coulombs)
Thus, +2.1 µC = +2.1 x 10^(-6) C
Now, using the equation:
charge = number of electrons × elementary charge
We can rearrange the equation to solve for the number of electrons:
number of electrons = charge / elementary charge
Substituting the given values:
number of electrons = (+2.1 x 10^(-6) C) / (1.6 x 10^(-19) C)
Performing the calculation:
number of electrons = 1.31 x 10^(13) electrons
Therefore, approximately 1.31 x 10^(13) electrons must be removed from an electrically neutral silver dollar to give it a charge of +2.1 µC.