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A parallel-plate air capacitor has a capacitance of 4 muF. What will be its new capacitance if (i) the distance between the plates is redced to half the initial distance (ii) a slab of dielectric constant 5 is introduced filling the entire space between the two plates? |
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Answer» SOLUTION :Data: `C_(1)=4 muF, d_(2) = d_(1)/2, k_(1)=1` (air), `k_(2)=5` `C=(Akepsilon_(0))/d` (i) With air as the dielectric, `C_(1) =(Ak_(1)epsilon_(0))/(d_(1))` and `C_(2) = (Ak_(1)epsilon_(0))/(d_(2))therefore C_(2)/C_(1) = d_(1)/d_(2)` `therefore` The new capacitance when the plate separation is reduced. `C_(2)= C_(1).d_(1)/d_(2) = 4 xx 2 =8 muF` (ii) With the plates separation `d=d_(1)`, `C_(1) = (Ak_(1)epsilon_(0))/(d_(1))` and `C_(2) = (Ak_(2)epsilon_(0))/d_(1)therefore C_(2)^(')/C_(1)therefore (C_(2)^('))/(C_(1)) = k_(2)/k_(1)` `therefore` the new capacitance when a dielectric slab is introduced between the plates, `C_(2)^(') =C_(1).k_(2)/k_(1) = 4 xx 5= 20 muF` |
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