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.

Explanation:

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

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

Putting all together:

Which for our values is:

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 is enough to describe an area).

answer: decibel

explanation: