Physics, asked by Anonymous, 10 months ago

if the relative permittivity of water is 80 then derive the relationship between Fwater and Fvaccum.what can be concluded from it?​

Answers

Answered by supratickghosh
2

Answer:

Fwater=81F air

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Answered by sreejamiyami
0

Answer:

Explanation:

Permittivity is one of the fundamental material parameters that affect the propagation of Electric Fields. Permittivity, typically denoted by the symbol ɛ (lowercase Epsilon), is a measure of how much the molecules of a material oppose an external electric field. The relative permittivity of a material compared to the permittivity of a vacuum (no electrons present) is known as the Dielectric Constant of the material.

The permittivity of free space (a vacuum) is a physical constant equal to approximately 8.85 x10−12 farad per meter (F/m).By definition,

permittivity of water = dielectric constant x permittivity of vacuum

Therefore,

permittivity of water = 80.10 x 8.85 x10–12 = 7.08885 x 10–10 farad/metre

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