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The answer depends on the vector field whose flux you are talking about. There are two vector fields that describe the electric part of electromagnetism: the electric field vector $\vec{E}$ and the electric displacement $\vec{D}$ (see Wiki page) - don't get too caught up in the name of the latter: it was named by James Clerk Maxwell (I believe) and its name ...


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The confusion is what you mean by $Q$. Flux is still $$\frac{q_{in}}{\epsilon_0}$$ But, the numerator has changed. If you take into account the negative charge $$-q(1-\frac 1 K)$$ induced, you will reach the result : $$\frac{q}{K\epsilon_0}$$ Remember, this $q$ is only the charge $q$, not accounting any other induced charges. It is denoted by ...


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In theory, provided that the surface area of the electrode remains constant and that the electrode material is conductive, the capacitance will not differ significantly when using different metals. This is just a consequence of the fact that for parallel plates, the capacitance only really depends on plate surface area. However, for modern electrochemical ...


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You can speak of a dielectric constant when the dielectric function is approximately constant for the given range of interest. Some materials, like air, have a slowly varying function with the wavelength; in that case, you can consider it constant. Also, if your problem involves a fixed wavelength (say, you consider the influence of a laser on a material), ...


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When one reads a statement, " dielectric constant of SiO2 is 5" It is the response of SiO2 to a dc electric field, which is normally measured like a capacitor. But then you probe the material with light (varying electric field) We get the ac response, like an elipsometery measurement



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