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A charge $q$ is enclosed by a closed surface immersed in a medium of relative permittivity (dielectric constant) $\varepsilon_r$. The total electric flux through the surface is
A$\dfrac{q}{\varepsilon_0}$
B$\dfrac{q\,\varepsilon_r}{\varepsilon_0}$
C$\dfrac{q}{4\pi\varepsilon_0}$
D$\dfrac{q}{\varepsilon_r\varepsilon_0}$
Answer & Solution
Correct answer: D. $\dfrac{q}{\varepsilon_r\varepsilon_0}$
1. In a medium the permittivity becomes $\varepsilon = \varepsilon_r\varepsilon_0$.
2. Gauss's law then reads $\Phi = \dfrac{q}{\varepsilon_r\varepsilon_0}$.
3. The medium reduces the flux by the factor $\varepsilon_r$ (e.g. water, $\varepsilon_r \approx 80$).
_Source: Samacheer Kalvi Class 12 Physics Vol.1, Unit 1 'Electrostatics', §1.6 / Evaluation Q.6_
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