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A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 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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Answer» Solution :Capacitance of capacitor with air as MEDIUM between the plates , `C_(0)=(in_(0)A)/(d)` `8 xx10^(-12)= (in_(0)A)/(d)` Now, dielectric fHled between two plates of capacitor and the distance between the PLATE is REDUCED to half, then the capacitance will be, `C = (in_(0)A)/((d)/(2)) = (2K in_(0)A)/(d)` `:. C= 2 xx6xx8xx10^(-12) ` ( From equation (1)) `=96 xx10^(-12) F ` `:.C = 96p E ` |
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