Q146 / 160 · AP EAMCET 2015 · Physics

If the dielectric constant of a substance is $K=\frac{4}{3}$, then the electric susceptibility $\psi_{\mathrm{e}}$ is

  • Correct answer
    A
    $\frac{\varepsilon_0}{3}$
  • B
    $3\ \varepsilon_0$
  • C
    $\frac{4}{3} \varepsilon_0$
  • D
    $\frac{3}{4} \varepsilon_0$

Solution

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, $\chi_e=\frac{\varepsilon_0}{3}$

Hence, the correct answer is option: 'A'.

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