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If the dielectric constant of a substance $K=\frac{4}{3}$, then the electric susceptibility $\chi$ in terms of vacuum permittivity $\varepsilon_0$ is
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The correct answer is:
$\frac{\varepsilon_0}{3}$
Given, $K=\frac{4}{3}$
For a dielectric, the electric displacement,
$\mathbf{D}=\varepsilon_0 \mathbf{E}+\mathbf{P} \quad \text { or } \quad D=\varepsilon_0 E+P$
$K \varepsilon_0 E=\varepsilon_0 E+\chi E \Rightarrow \chi=(K-1) \varepsilon_0$
$\Rightarrow \quad \chi=\varepsilon_0\left(\frac{4}{3}-1\right) \Rightarrow \chi=\frac{\varepsilon_0}{3}$
For a dielectric, the electric displacement,
$\mathbf{D}=\varepsilon_0 \mathbf{E}+\mathbf{P} \quad \text { or } \quad D=\varepsilon_0 E+P$
$K \varepsilon_0 E=\varepsilon_0 E+\chi E \Rightarrow \chi=(K-1) \varepsilon_0$
$\Rightarrow \quad \chi=\varepsilon_0\left(\frac{4}{3}-1\right) \Rightarrow \chi=\frac{\varepsilon_0}{3}$
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