Answer:
[tex]8.89\cdot 10^4 V/m[/tex]
Explanation:
The electric field strength between the plates of a parallel plate capacitor is given by
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where
[tex]\sigma[/tex] is the surface charge density
[tex]\epsilon_0[/tex] is the vacuum permittivity
Here we have
[tex]A=3.00 \cdot 3.00 = 9.00 cm^2 = 9.0\cdot 10^{-4} m^2[/tex] is the area of the plates
[tex]Q=0.708 nC = 0.708 \cdot 10^{-9} C[/tex] is the charge on each plate
So the surface charge density is
[tex]\sigma=\frac{Q}{A}=\frac{0.708\cdot 10^{-9}}{9.0\cdot 10^{-4} m^2}=7.87\cdot 10^{-7} C/m^2[/tex]
And now we can find the electric field strength
[tex]E=\frac{\sigma}{\epsilon_0}=\frac{7.87\cdot 10^{-7}}{8.85\cdot 10^{-12}}=8.89\cdot 10^4 V/m[/tex]
