1.

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.

Answer»

the dielectric constant for the dielectric SLAB is 4
without the dielectric, the capacitance of the capacitor would be `360muF`
the PLATE area is `60L^(2)`
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`

SOLUTION :`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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