Answer:
[tex]5.5\cdot 10^{-11} C[/tex]
Explanation:
The capacitance of the parallel-plate capacitor is given by:
[tex]C=\epsilon_0 \frac{A}{d}[/tex]
where
[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity
[tex]A=190 mm^2 = 190 \cdot 10^{-6} m^2[/tex] is the area of the plates
[tex]d=1.2 mm = 0.0012 m[/tex] is the separation between the plates
Substituting,
[tex]C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F[/tex]
The energy stored in the capacitor is given by
[tex]U=\frac{Q^2}{2C}[/tex]
Since we know the energy
[tex]U=1.1 nJ = 1.1 \cdot 10^{-9} J[/tex]
we can re-arrange the formula to find the charge, Q:
[tex]Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C[/tex]
