A parallel plate capacitor with air between the plates has a capacitance of 8pF (1 pF = 10^(-12)F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ?

A parallel plate capacitor with air between the plates has a capacitan

1.	A parallel plate capacitor with air between the plates has a capacitance of 8pF (1 pF = 10^(-12)F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ?
Answer» Solution :Here `C = 8 pF = (epsi_0A)/d` Now `d. = d/2` and space is filled with a SUBSTANCE having K = 6. HENCE, new capacitance `C.= (K epsi_0A)/(d.) = (6epsi_0A)/((d/2)) = 12 (epsi_0A)/d = 12 xx 8 pF-96 pF.`

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A parallel plate capacitor with air between the plates has a capacitance of 8pF (1 pF = 10^(-12)F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ?

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