Just a simple question regarding $\epsilon$, the absolute permittivity. I read that it measures the resistance to of certain medium to 'permit' the formation of an electrical field. Does this mean that given a electrical field $E$, one needs more energy to produce it in a medium with a higher $\epsilon$ ? Thanks in advance.
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$\begingroup$ The energy density stored in a field distribution is $\frac{1}{2}\epsilon|E|^2$. So fixing $|E|$, you get your result. $\endgroup$– Chris GerigCommented Jan 14, 2014 at 5:59
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If you need to set up a field of magnitude $E$ in a dielectric medium of relative permittivity $\epsilon_r$ then you need to originally supply an electric field of $\epsilon_r E$, Now since a larger field would be provided with more charge, and bringing new charge to an existing charge configuration would require energy, definitely more energy is required to set up an electric field in presense of a dielectric medium than it would be in perfect vacuum.
answered Jan 14, 2014 at 9:13