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A parallel plate capacitor with only air between the plates has a capacitance of 8pF. 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 :`C_0 = (A epsi_0)/(d) = 8pF` When the distance is reduced to half and dielectric MEDIUM FILLS the GAP, the new capacitance will be `C = (epsi_r Aepsi_0)/(d//2) = (2 epsi_r A epsi_0)/(d)` ` = 2 epsi_r C_0` `C = 2 xx 6 xx 8 = 96 PF` |
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