TS EAMCET · Physics · Electrostatics
If the dielectric constant of a substance \(K=\frac{4}{3}\), then the electric susceptibility \(\chi\) in terms of vacuum permittivity \(\varepsilon_0\) is
- A \(\frac{\varepsilon_0}{3}\)
- B \(3 \varepsilon_0\)
- C \(\frac{4}{3} \varepsilon_0\)
- D \(\frac{3}{4} \varepsilon_0\)
Answer & Solution
Correct Answer
(A) \(\frac{\varepsilon_0}{3}\)
Step-by-step Solution
Detailed explanation
Given, \(K=\frac{4}{3}\) For a dielectric, the electric displacement, \(\mathbf{D}=\varepsilon_0 \mathbf{E}+\mathbf{P} \quad \text { or } \quad D=\varepsilon_0 E+P\) \(K \varepsilon_0 E=\varepsilon_0 E+\chi E \Rightarrow \chi=(K-1) \varepsilon_0\)…
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