Answer
$q\prime==4.1nC$
Work Step by Step
The magnitude of the induced surface charge, $q\prime$, can be calculated as
$q\prime=q-\frac{q}{k}$
We plug in the known values to obtain:
$q\prime=5\times10^{-9}-\frac{5\times10^{-9}}{5.4}$
$q\prime=\frac{5.4(5\times10^{-9})-5\times10^{-9}}{5.4}$
$q\prime=\frac{5\times10^{-9}(5.4-1)}{5.4}$
$q\prime=\frac{5\times10^{-9}(4.4)}{5.4}=4.1\times10^{-9}$
$q\prime==4.1nC$
