Answer:
[tex]5.63\cdot 10^{-6} C[/tex]
Explanation:
The capacitor of a parallel-plate capacitor is given by:
[tex]C=\epsilon_0 \frac{A}{d}[/tex]
where
A is the area of each plate
d is the separation between the plates
[tex]\epsilon_0[/tex] is the vacuum permittivity
The energy stored in a capacitor instead is given by
[tex]U=\frac{1}{2}\frac{Q^2}{C}[/tex]
where
Q is the charge stored in each plate
Substituting the expression we found for C inside the last formula,
[tex]U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}[/tex]
And re-arranging it
[tex]Q=\sqrt{\frac{2U\epsilon_0 A}{d}}[/tex]
Now if we substitute
[tex]d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J[/tex]
We find the charge stored on the capacitor:
[tex]Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C[/tex]
