Answer:
[tex]9.5 \cdot 10^{-8} C[/tex]
Explanation:
The electric field between two charged plates is uniform, and its strength is given by
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where
[tex]\sigma=\frac{Q}{A}[/tex] is the charge density, with Q being the charge on one plate and A the area of the plate
[tex]\epsilon_0=8.85 \cdot 10^{-12} C^ N^{-1} m^{-2}[/tex] is the dielectric constant
Since we know the magnitude of the field, [tex]E=6.8 \cdot 10^6 N/C[/tex], we can find the charge density:
[tex]\sigma=\epsilon_0 E=(8.85 \cdot 10^{-12})(6.8 \cdot 10^6 )=6.0 \cdot 10^{-5} C m^{-2}[/tex]
The area of each plate is
[tex]A=(0.04 m)^2=1.6 \cdot 10^{-3} m^2[/tex]
So the charge on each electrode is
[tex]Q=\sigma A=(6.0 \cdot 10^{-5})(1.6 \cdot 10^{-3})=9.5 \cdot 10^{-8} C[/tex]
