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If the dielectric constant of a substance is $K=\frac{4}{3}$, then the electric susceptibility $\psi_{\mathrm{e}}$ is
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Verified Answer
The correct answer is:
$\frac{\varepsilon_0}{3}$
The dielectric constant permittivity and susceptibility are relative as,
$\begin{aligned}
& K=1+\frac{\chi_e}{\varepsilon_0} \Rightarrow \frac{4}{3}=1+\frac{\chi_e}{\varepsilon_0} \\
& \Rightarrow \chi_e=\varepsilon_0\left(\frac{4}{3}-1\right) \\
& \Rightarrow \quad \chi_e=\frac{\varepsilon_0}{3}
\end{aligned}$
The electric susceptibility, $\chi_e=\frac{\varepsilon_0}{3}$
$\begin{aligned}
& K=1+\frac{\chi_e}{\varepsilon_0} \Rightarrow \frac{4}{3}=1+\frac{\chi_e}{\varepsilon_0} \\
& \Rightarrow \chi_e=\varepsilon_0\left(\frac{4}{3}-1\right) \\
& \Rightarrow \quad \chi_e=\frac{\varepsilon_0}{3}
\end{aligned}$
The electric susceptibility, $\chi_e=\frac{\varepsilon_0}{3}$
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