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Answered 2008-10-25 13:23:29

From the formula, E = V/d, where V is the voltage and d is the distance, it can be seen that the electric field and the distance are inversely related. Thus, as the distance between the parallel plate capacitors is reduced to half, the electric field is increased twice. Moreover, is a dielectric constant k is introduced, the capacitance will increase. This direct relationship can be seen in the formula, C = [k(Єo)A]/d, where k is the dielectric constant. The lowest possible value of k is 1; and that is when the dielectric is a vacuum. Other dielectric constants are greater than 1, such as Teflon which has a dielectric value of 2.1 As the capacitance increases, the electric field also increases. (E = 1/2(CV^2))

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What is the capacitance of a parallel plate capacitor with a dielectric of oil-soaked paper The dielectric constant is 3.8 the distance between the plates is 0.00625 inches and the plate area is 0.25?

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