Answer
$q=315mC$
Work Step by Step
We know that:
$q=CV$
We plug in the known values to obtain:
$q=(25\times 10^{-6})(4200)=0.105C$
Now, the total charge can be determined as:
$q_{total}=3q$
We plug in the known values to obtain:
$q=3(0.105)=0.315=315\times 10^{-3}=315mC$
