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A parallel plate capacitor is to be constructed which can store q=10μC charge at V = 1000 volt. The minimum plate area of the capacitor is required to be A1 when space between the plates has air. If a dielectric of constant K = 3 is used between the plates, the minimum plate area required to make such a capacitor is A2. The breakdown field for the dielectric is 8 times that of air. Find A1A2.

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answer is 24.

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Detailed Solution

For area of the plate to be minimum, the separation between the plates must be minimum. The separation between the plates is limited by the breakdown strength of the dielectric.For air capacitor Vdmin=Eair Eair = Breakdown field for air ∴ dmin=VEair Now ε0Amindmin=C⇒ Amin=Cε0VEair⇒ A1=CVε0Eair With dielectric, similar calculation gives, A2=CVKε0Edielect ∴ A1A2=KEdielec Eair =3×8=24

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A parallel plate capacitor is to be constructed which can store q=10μC charge at V = 1000 volt. The minimum plate area of the capacitor is required to be A1 when space between the plates has air. If a dielectric of constant K = 3 is used between the plates, the minimum plate area required to make such a capacitor is A2. The breakdown field for the dielectric is 8 times that of air. Find A1A2.