Question

A 3.0 pF capacitor consists of two parallel plates that have surface charge densities of 1.0 nC/mm2 . If the potential between the plates is 27.0 kV, find the surface area of one of the plates.

Answer 1

Answer:

$A=81mm^2$

Explanation:

We know that for a capacitor $Q=CV$, where Q is the charge of one plate, C the capacitance and V the potential between the plates.

We also know that $Q=\sigma A$, since $\sigma$ is the surface charge density and A the area of the plate (both equal in our case).

Putting all together:

$A=\frac{CV}{\sigma}$

Which for our values is:

$A=\frac{(3\times10^{-12}F)(27\times10^3V)}{1\times10^{-9}C/mm^2}=81mm^2$

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 $mm^2$ is enough to describe an area).