Answer:
The surface charge density on the plate, [tex]\sigma=7.5\times 10^{-5}\ C/m^2[/tex]
Explanation:
It is given that,
Area of parallel plate capacitor, [tex]A=2\times 10^{-3}\ m^2[/tex]
Separation between the plates, [tex]d=5\times 10^{-2}\ mm[/tex]
Electric field between the plates, [tex]E=8.5\times 10^{6}\ V/m[/tex]
We need to find the surface charge density on the plates. The formula for electric field is given by :
[tex]E=\dfrac{\sigma}{\epsilon}[/tex]
Where
[tex]\sigma[/tex] = surface charge density
[tex]\sigma=E\times \epsilon[/tex]
[tex]\sigma=8.5\times 10^{6}\ V/m\times 8.85\times 10^{-12}\ C^2/Nm^2[/tex]
[tex]\sigma=0.000075\ C/m^2[/tex]
[tex]\sigma=7.5\times 10^{-5}\ C/m^2[/tex]
Hence, this is the required solution.
