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A parallel plate capacitor of capacitance 'C' is charged to 'V' volt by a battery . After some time the battery is disconnected and the distance between the plates is doubled . Now a slab of dielectric constant 1 lt k lt 2 is introduced to fill the space between the plates . How will the following be affected ? (i) The electric field between the plates of the capacitor . (ii) The energy stored in the capacitor . Justify your answer in each case. |
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Answer» SOLUTION :Let initially a capacitor of CAPACITANCE C is charged to a potential difference of V volt so that charge on the capacitor Q = CV On disconnecting the battery , charge Q on capacitor remainsconserved . Now distance between the plates of capacitor is DOUBLED i.e., d. =2 d and the intervening space is filled with a dielectric medium for which `1 lt k lt 2 `. Thus , new capacitance `C. = (K in_(0) A)/(d.) = (K in_(0) A)/(2d) = (K)/(2) C` . As `Klt 2 ` . hence `C. lt C` (i) Initial value of electric field was `F = (sigma)/(in_(0)) = (Q)/(A in_(0))` and new value of electric field `E. = (sigma)/(K in_0) = (Q)/(A K in_(0)) = (E)/(K)` . So the electric field between the plates of the capacitor . (ii) Initial value of energy stored in capacitor `U = (Q^2)/(2C)` and new value of energy stored in capacitor `U. = (Q^2)/(2C.)` As `C. lt C` , hence `U. gt U` . So the value of energy stored in the capacitor increases . 
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