Answer:
[tex]A=81mm^2[/tex]
Explanation:
We know that for a capacitor [tex]Q=CV[/tex], where Q is the charge of one plate, C the capacitance and V the potential between the plates.
We also know that [tex]Q=\sigma A[/tex], since [tex]\sigma[/tex] is the surface charge density and A the area of the plate (both equal in our case).
Putting all together:
[tex]A=\frac{CV}{\sigma}[/tex]
Which for our values is:
[tex]A=\frac{(3\times10^{-12}F)(27\times10^3V)}{1\times10^{-9}C/mm^2}=81mm^2[/tex]
Where we notice that the S.I. units combination FV/C must not have units (we can verify it directly from their definitions or we notice that [tex]mm^2[/tex] is enough to describe an area).
