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poizon [28]
3 years ago
10

A 3.0 pF capacitor consists of two parallel plates that have surface charge densities of 1.0 nC/mm2 . If the potential between t

he plates is 27.0 kV, find the surface area of one of the plates.
Physics
1 answer:
Archy [21]3 years ago
8 0

Answer:

A=81mm^2

Explanation:

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

We also know that Q=\sigma A, since \sigma is the surface charge density and <em>A</em> 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).

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Answer:

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