Answer
The total charge on the two capacitors is $~~3.6\times 10^{-12}~C$
Work Step by Step
We can write an expression for the capacitance:
$C = \frac{\epsilon_0~A}{d}$
We can write an expression for the potential difference:
$V = E~d$
We can write an expression for the charge on a capacitor:
$q = CV = (\frac{\epsilon_0~A}{d})(E~d) = \epsilon_0~A~E$
We can find the total charge on the two capacitors:
$q = \epsilon_0~A_1~E_1+\epsilon_0~A_2~E_2$
$q = \epsilon_0~(A_1~E_1+A_2~E_2)$
$q = (8.854\times 10^{-12}~C/V~m)~[(2000~V/m)(1.5\times 10^{-4}~m^2)+(1500~V/m)(0.70\times 10^{-4}~m^2)]$
$q = 3.6\times 10^{-12}~C$
The total charge on the two capacitors is $~~3.6\times 10^{-12}~C$.
