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Fig. shows two identical capacitorsC_(1) and C_(2) each of 1 muF capacitance, connected to a batteryof 6VInitially,swich S is closed. Aftersometime,S is leftopen and dielectricslabs of dielectricconstant K = 3 are insteredto fillcompelelty the space betweenthe plates of two capacitors. How will the (i) charge and (ii) potential difference between the platesof the capacitors be affectedafter teh slabs are inserted ? |
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Answer» Solution :In Fig, when switch S is closed, both`C_(1) and C_(2)` are chargedto 6Vpotential, i.e, `V_(1) = 6V and V_(2) = 6V` when S is left open, `C_(1)` is still connectedto battery. On introducingslabof DIELECTRICCONSTANT. K = 3. `C'_(1) = KC_(1) = 3mu F= 3xx10^(-6) F`, `V'_(1) = V_(1) = 6V` `Q_(1) = C'_(1) V'_(1) = 3xx10^(-6) xx6 = 18xx10^(-6)C`. However `C_(2)` is disconnecedfrom battery now, when S is left open. Due to introducedof slab, `C'_(2) = KC_(2) = 3xx10^(-6)F` `Q'_(2) = Q_(2) = C_(2) V_(2) = 1XX10^(-6)xx6 = 6xx10^(-6) C` `:. V_(2) = (Q_(2))/(C_(2)) = (6xx10^(6))/(3xx10^(-6)) = 2V` |
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