Answer:
9.1 * 10²⁴ electron
Explanation:
The total charge is distributed over the two objects:
[tex]Q_{total}/2=(2.3*10^{-6}C+2.4 * 10^-^6C)/2=2.35 * 10^-^6C\\[/tex]
The plate and the rod must have [tex]Q_{total}/2\\[/tex].
So the charge transferred from the plate to the rod is:
[tex]Q_{transfered}=3.8*10^{-6}C-Q_{total}/2\\\\=3.8*10^{-6}C- 2.35 *10^{-6} C\\\\=1.45 *10^{-6}C\\\\[/tex]
[tex]N_{electrons}=Q_{transfered}/q_{electron}\\\\= 1.45 * 10^-^6 C/(-1.6*10^{-19}C)\\\\=9.1*10^{24}electrons\\\\\\[/tex]
The number of electrons (n) that must be transferred from the plate to the rod is [tex]9.1 \times 10^{-19}\;electrons[/tex].
Given the following data:
Scientific data:
To calculate the number of electrons (n) that must be transferred from the plate to the rod:
For the plate and rod to have the same charge, the number of electrons (n) flowing through them must be equal to half the total quantity of charges possessed by both of them.
[tex]Q_t =\frac{Q_1 + Q_2}{2} \\\\Q_t =\frac{2.3 \times 10^{-6} + 2.4 \times 10^{-6} }{2} \\\\Q_t=\frac{4.7 \times 10^{-6} }{2} \\\\Q_t=2.35\times 10^{-6} \;C[/tex]
Next, we would calculate the quantity of charge that was transferred from the plate to the rod:
[tex]Q_r=3.6 \times 10^{-6}- 2.35 \times 10^{-6}\\\\Q_r = 1.45 \times 10^{-6}\;C[/tex]
For the number of electrons (n):
[tex]n=\frac{Q_r}{q} \\\\n=\frac{1.45 \times 10^{-6}}{1.6 \times 10^{-19}} \\\\n=9.1 \times 10^{-19}\;electrons[/tex]
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