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What is the effect of dielectric medium in comparison to vacuum on electric field and electric flux?

permittivity relates to a material's ability to resist an electric field (while, unfortunately, the word stem "permit" suggests the inverse quantity).

Source: Permittivity Wikipedia

Therefore, higher dielectric constant means higher relative permittivity which means more resistance to the electric field.

But a higher dielectric constant means an increase in electric flux which leads to a higher capacitance!

Please explain the resistance to the electric field with an increase in electric flux?

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Dielectric constant $K$ is actually the same thing as relative permittivity, and it increases the overall permittivity $\epsilon$. So in general, whenever you see the permittivity of free space $\epsilon_0$ in an equation, if you're dealing with a dielectric, you can multiply it by the dielectric constant and see how the equation changes.

For example, since $C = \epsilon_0 \frac{A}{D}$, multiplying $\epsilon_0$ by $K$ increases capacitance by $K$. Since $\oint \mathbf E \cdot \mathrm d \mathbf A = \frac{Q}{\epsilon_0}$ by Gauss's Law, multiplying $\epsilon_0$ by $K$ decreases electric field magnitude by $K$.

There is no contradiction because there is no increase in electric flux. The dielectric decreases the electric field magnitude, which decreases the electric flux and decreases the voltage across a capacitor as well. $C = \frac{Q}{V}$ and the capacitance increases because the voltage decreases while the charge remains the same.

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  • $\begingroup$ excuse me, could you give me a justification or source of this hint. (I mean, change $\varepsilon_0$ for $K\varepsilon_0$?). $\endgroup$ – Ragnar1204 Apr 8 '18 at 9:24

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