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A dielectric slab fills the space between the plates of a parallel-plate capacitor. The magnitude of the bound charge on the slab is `75% of the magnitude of the free charge on the plates. The capacitance is `480muf` and the maximum charge that can be stored on the capacitor is `240varepsilon_(0)L^(2) E_(max)` where `E_(max)` is the breakdown field.A. the dielectric constant for the dielectric slab is 4B. without the dielectric, the capacitance of the capacitor would be `360muF`C. the plate area is `60L^(2)`D. If the dielectric slab is having the same area as the capacitor plate but the width half that of the capacitor, the capacitance would be `192muF` |
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Answer» Correct Answer - A::C::D `Q/(Ain_(0))=E_(max)` `(240in_(0)L^(2))/(kAin_(0)) A=60L^(2)` `(4in_(0)A)/d=480` `(in_(0)A)/(d-d/2(1-1/4))=120xx8/5=192` |
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