Yungnaudie
07.06.2020 •
Physics
a. When the electric field between the plates is 75% of the dielectric strength and energy density of the stored energy is 2800 J/m3, what is the value of the dielectric constant k?
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Ответ:
The value of the dielectric constant k = 1.8
Explanation:
If C= ε A/d and
Electrostatic energy W = 1/2CV^2
Substitutes C in the first formula into the energy formula.
W = 1/2 ε A/d × V^2
Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed
Substitute V = Ed into the energy W.
W = 1/2 × ε A/d ×( Ed )^2
W = 1/2 × ε A/d × E^2 × d^2
d will cancel one of the ds
W = 1/2 × ε Ad × E^2
W/Ad = 1/2 × ε × E^2
W/V = 1/2 × ε E^2
Where Ad = volume V
E = dielectric strength
εo = permittivity of free space = 8.84 x 10^-12 F/m
W/V = 2800 J/m^3
Let first calculate the dielectric strength
2800 = 1/2 × 8.84×10^-12 × E^2
5600 = 8.84×10^-12E^2
E^2 = 5600/8.84×10^-12
E = sqrt( 6.3 × 10^14)
E = 25 × 10^7
75% of E = 18.9 × 10^6Jm
The permittivity of the material will be achieved by using the same formula
2800 = 1/2 × ε E^2
2800 = 0.5 × ε × (18.9×10^6)^2
2800 = ε × 1.78 × 10^14
ε = 2800/1.78×10^14
ε = 1.57 × 10^-11
Dielectric constant k = relative permittivity
Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is
k = 1.57×10^-11/8.84×10^-12
k = 1.776
k = 1.8 approximately
Therefore, the value of the dielectric constant k is 1.8
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