Question

A plate carries a charge of 3.8 UC, while a rod carries a charge of 1.9 C. How many electrons must be transferred from the plate to the rod, so that both objects have the same charge?

Answer 1

Answer:

$N_{electrons}=Q_{transfered}/q_{electron}=5.94*10^{18}electrons$

Explanation:

The total charge is distributed over the two objects:

$Q_{total}/2=(3.8*10^{-6}C+1.9C)/2=0.9500019C$

The plate and the rod must have $Q_{total}/2$. So the charge transferred from the plate to the rod is:

$Q_{transfered}=3.8*10^{-6}C-Q_{total}/2=3.8*10^{-6}C-0.9500019C=-0.9499981C$

Number of electrons:

$N_{electrons}=Q_{transfered}/q_{electron}=-0.9499981C/(-1.6*10^{-19}C)=5.94*10^{18}electrons$