The electric field induced in a dielectric when placed in an external field is 1/10 times the external field. Calculate the relative permittivity of the dielectric.
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Relative permittivity is the ratio of permittivity of medium or permittivity of vacuum. e.g., ∈ₐ = ∈/∈₀ here ∈ₐ shows relative permittivity of the dielectric.
It is also known as dielectric constant e.g., K
Again, Electric field intensity , E = Q/4π∈₀Kr² in a medium of dielectric constant is K
Now, according to question,
Electric field intensity induced in a dielectric when placed in an external field is 1/10 times of external field.
Means , Electric in dielectric = Electric field in vacuum/10
Q/4π∈₀Kr² = Q/4π∈₀r²10
K = 10
Hence, relative permittivity of dielectric is 10
It is also known as dielectric constant e.g., K
Again, Electric field intensity , E = Q/4π∈₀Kr² in a medium of dielectric constant is K
Now, according to question,
Electric field intensity induced in a dielectric when placed in an external field is 1/10 times of external field.
Means , Electric in dielectric = Electric field in vacuum/10
Q/4π∈₀Kr² = Q/4π∈₀r²10
K = 10
Hence, relative permittivity of dielectric is 10
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6
Answer:
Relative permittivity ∈r is defined as the the ratio of absolute permittivity ∈ to the permittivity of free space ∈o
∈r=∈/∈o which is equal to the dielectric constant K And
as given E=10Eo∈r=10
Explanation:
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