InterviewSolution
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InterviewSolution
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Inside a parallel-plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to `eta = 0.60` of the gap width. When the plates is absent the capacitor cpaacitance equals `c = 20 muF` . First, the capacitor was connected in parallel to a constant voltage source producting `V = 200 V`, then it was disconnected for it, after which the plates was slowly removed from teh gap. FInd the work perfomed during the removel, if the plate is (a) made of metal, (b) made of glass. |
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Answer» (a) When metal plate of thickness `eta d` is inserted inside the capacitor, capacitance of the system becomes `C_(0) = (epsilon_(0) S)/(d (1 - eta))` Now, intially charge on the capacitor, `q_(0) = C_(0) V = (epsilon_(0) S V)/(d (1 - eta))` Finally, capacitance of the capacitor, `C = (epsilon_(0) S)/(d)` As the source is disconnected, charge on the plates will remain same during the process. Now, from energy conservation, `U_(f) - U_(i) = A_("agent")` (as cell does no work) or, `(1)/(2) (q_(0)^(2))/(C) - (1)/(2) (q_(0)^(2))/(C_(0)) = A_("agent")` Hence `A_("agent") = (1)/(2) [(epsilon_(0) S V)/(d (1 - eta))]^(2) [(1)/(C) - ((1 - eta))/(C)] = (1)/(2) (C V^(2) eta)/((1 - eta)^(2)) = 1.5 mJ` (b) Initally, capacitance of the system is given by : `C_(0) = (C epsilon)/(eta (1 - epsilon) + epsilon)` (this is the capacitance of two capacitors in series) So, charge on the plate , `q_(0) = C_(0) V` Capacitance of the capacitor, after the glass plate has been removed equals `C` from energy conservation, `A_("agent") = U_(f) - U_(i)` `= (1)/(2) q_(0)^(2) [(1)/(C) - (1)/(C_(0))] = (1)/(2) (C V^(2) epsilon eta (epsilon - 1))/([epsilon - eta (epsilon -1)]^(2)) = 0.8 mJ` |
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