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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Solution

Relative permittivity $\in_{r}$ is defined as the the ratio of absolute permittivity $\in$ to the permittivity of free space $\in_{o}$

$\in_{r} = \frac{\in}{\in_{o}} w h i c h i s e q u a l t o t h e d i e l e c t r i c c o n s \tan t K A n d K = \frac{E_{o}}{E} w h e r e E$

$S o \in_{r} = \frac{E_{0}}{E} A n d a s g i v e n E = \frac{1}{10} E_{o} \in_{r} = 10$

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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.

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 K=EoE where E So ∈r=E0EAnd as given E=110Eo∈r=10