A dielectric of thickness 5 cm and a dielectric constant 10 is introduced between the plates of a parallel plate capacitor having plate area 500 cm and separation between the plates 10 . Calculate the capacitance of the capacitor with the dielectric slab.
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Step 1
Given: Plate area (A) = 500 cm^2 = 500 * 10^-4 m^2 Separation between plates (d) = 10 cm = 0.1 m Thickness of dielectric (t) = 5 cm = 0.05 m Dielectric constant (K) = 10 The effective capacitance (C') when the dielectric slab is introduced is given by: C' = (K * Show more…
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