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In a parallel plate air capacitor the distance between plates is reduced to one fourth and the space between them is filled with a dielectric medium of constant 2. If the initial capacity of the capacitor is `4muF`. then its new capacity is .A. `32 muF`B. `18 muF`C. `8 muF`D. `44 muF` |
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Answer» Correct Answer - A The capacitance of parallel plate air capacitor is given by `C_(0) = (epsi_(0)A)/(d)" "……..(i)` Where `,epsi_(0)` = permittivity of the medium. A = area of plates and d= distance between the plates When the distance between plates in reduced and a dielectric slab is introduced , then capacitance becomes . `C= (k epsi_(0)A)/(d_(1)) " "........(ii)` where , k = dielectric constant of medium . Here`k= 2, d_(1) = (d)/(4) and c_(0) = 4 mu F = 4 xx 10^(-6)F` From Eq.(ii) we get `C = 4 ((k epsi_(0) A)/(d_(1))) = 4k((epsi_(0)A)/(d))` ` = 4kC_(0)" "....(iii)` [from Eq.(i)] Substituting given values in Eq. (iii) we , get `C = 4xx 2 xx 4 = 32 muF` |
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