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Question
The two square faces of a rectangular dielectric slab (dielectric constant 4⋅0) of dimensions 20 cm × 20 cm × 1⋅0 mm are metal-coated. Find the capacitance between the coated surfaces.
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Solution
The area of the plates of the capacitor is given by
`A = 20 "cm" xx 20 "cm" = 400 "cm"^2`
⇒ `A = 4 xx 10^-2 "m"`
The separation between the parallel plates is given by
`d = 1 "m" = 1 xx 10^-3 "m"`
Here, the thickness of the dielectric is the same as the separation between the parallel plates.
Thus, the capacitance is given by
`C = (∈_0Ak)/d = ((8.85 xx 10^-12) xx (4 xx 10^-2) xx 4)/10^-3 = 1.42 "nF"`
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