Answer:
2.40 x 10⁻¹³ C
Explanation:
[tex]n_{e}[/tex] = number of electrons = 6.25 x 10⁶
[tex]q_{e}[/tex] = charge on electron = - 1.6 x 10⁻¹⁹ C
[tex]n_{p}[/tex] = number of protons = 7.75 x 10⁶
[tex]q_{p}[/tex] = charge on proton = 1.6 x 10⁻¹⁹ C
Net charge is given as
Q = [tex]n_{e}[/tex] [tex]q_{e}[/tex] + [tex]n_{p}[/tex] [tex]q_{p}[/tex]
Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)
Q = 2.40 x 10⁻¹³ C
The net charge on the system will be 2.40 x 10⁻¹³ C
It is given that,
[tex]n_{e}[/tex] = number of electrons = 6.25 x 10⁶
[tex]q_{e}[/tex] = charge on electron = - 1.6 x 10⁻¹⁹ C
[tex]n_{p}[/tex] = number of protons = 7.75 x 10⁶
[tex]q_{p}[/tex] = charge on proton = 1.6 x 10⁻¹⁹ C
The net charge is given as
[tex]Q=n_{e} q_{e} +n_{p}q_{p}[/tex]
[tex]Q= (-1.6\times10^{-19}) (6.25\times10^{6}) +(1.6\times10^{-19})(7.75\times10^{6} )[/tex]
[tex]Q=2.4\times10^{-13} C[/tex]
Thus net charge on the system will be [tex]Q=2.4\times10^{-13} C[/tex]
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