A parallel plate capacitor is to be constructed which can store $$q=10$$μC 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 $$A1A2$$.

Question

A parallel plate capacitor is to be constructed which can store $$q=10$$μC 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 $$A1A2$$.

Infinity Learn · Accepted Answer

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 ε$$0Amindmin=C$$⇒ $$Amin=C$$ε$$0VEair$$⇒ $$A1=CV$$ε$$0Eair$$ With dielectric, similar calculation gives, $$A2=CVK$$ε$$0Edielect$$ ∴ $$A1A2=KEdielec$$ Eair $$=3$$×$$8=24$$

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