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I just read it : "More electric flux exists in a medium with a high permittivity" I don't get this.

Field $\vec E$ is inversely proportional to permittivity $\epsilon$. So if $\epsilon$ is high then field should be small and hence the flux should also be small. Isn't it ?

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  • $\begingroup$ This is not correct. "More electric flux exists in a medium with a high permittivity" is correct when the electric field is held constant like in a capacitor connected to a voltage source. The electric flux is the surface integral of the dielectric displacement $\vec D=\epsilon_0 \epsilon_r\vec E$ which is the electric flux density. $\endgroup$
    – freecharly
    Commented Oct 14, 2016 at 3:49

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It's actually the other way. Flux density is inversely proportional to permittivity. As permittivity is by definition is the resistance offered by the medium to electric field, higher permittivity would only lower the flux.

Note: In this answer, "flux" is the flux of the electric field vector $\vec E$. The OP citation More electric flux exists in a medium with a high permittivity obviously relates to the flux of the electric displacement $\vec D$, which is often also called "electric flux" and used in Gauss' law.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Oct 19, 2016 at 20:47
  • $\begingroup$ So why it is named permittivity and not resistance? $\endgroup$ Commented Mar 9, 2021 at 16:25

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