Answer
$60\mu C$
Work Step by Step
We can find the initial charge of the capacitor as:
$q=CV$
We plug in the known values to obtain:
$q=6\times 10^{-6}\times 10=60\times 10^{-6}=60\mu C$
As we know that:
$C=\frac{A\epsilon_{\circ}}{d}$
The above equation shows that if the separation is halved then the capacitance will be doubled.
Thus the final charge of the capacitor is given as:
$q=(2C)(V)$
We plug in the known values to obtain:
$q=(2\times 6\times 10^{-6})(10)=120\times 10^{-6}=120\mu C$
Now the additional charge transmitted is
$120-60=60\mu C$
