Flux coming out from a positive charge of 8 | vedclass

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(B) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_m}$,where $\varepsi

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$\frac{2}{\varepsilon_0}$

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(B) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_m}$,where $\varepsilon_m$ is the permittivity of the medium.Given charge $q = 8 \ C$.The permittivity of the medium is $\varepsilon_m = K \varepsilon_0$,where $K$ is the dielectric constant and $\varepsilon_0$ is the permittivity of free space.Given $K = 4$,so $\varepsilon_m = 4 \varepsilon_0$.Substituting these values into the flux formula:$\phi = \frac{8}{4 \varepsilon_0} = \frac{2}{\varepsilon_0}$.

Flux coming out from a positive charge of $8 \ C$,placed in a medium of dielectric constant $4$ is

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