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A dielectric slab fills the space between the square plates of a parallel-plate capacitor. The side of each plate of the capacitor
\( \mathrm{P} \) is \( L \). The magnitude of the bound charges on the slab is \( 75 \% \) of W the magnitude of the free charge on the plates. The capacitance is \( 480 \mu \mathrm{F} \) and the maximum charge that can be stored on the capacitor is \( 240 \varepsilon_{0} L^{2} E_{\max } \), where \( E_{\max } \) is the breakdown strength of the medium :
(A) The dielectric constant for the dielectric slab is 4
(B) Without the dielectric, the capacitance of the capacitor would be \( 360 \mu F \)
(C) The plate area is \( 60 L^{2} \)
(D) If the dielectric slab is having the same area as that of the capacitor plate but the width is half the separation between plates of capacitor, the capacitance would be \( 192 \mu \mathrm{F} \).
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