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

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.

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niczorrrr 4 years 2021-08-31T01:17:43+00:00 1 Answers 137 views 0

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  1. Edana
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    2021-08-31T01:19:42+00:00
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    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).

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