Answer
(a) $2.06\times 10^{-13}C$
(b) $-7.5\times 10^{-18}C$
Work Step by Step
(a) We know that
$Q_{net}=N_pe+N_e(-e)=e(N_p-N_e)$
We plug in the known values to obtain:
$Q_{net}=(1.6\times 10^{-19})(7.44\times 10^6-6.15\times 10^6)=2.06\times 10^{-13}C$
(b) We know that
$Q_{net}=N_pe+N_e(-e)=e(N_p-N_e)$
We plug in the known values to obtain:
$Q_{net}=(1.6\times 10^{-19})(165- 212)=-7.5\times 10^{-18}C$
