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A parallel - plate capacitor, filled with a dielectric of dielectric constant k, is charged to a potential V_(0). It is now disconnected from the cell and the slab is removed. If it now discharges, with time constant tau, through a resistance, then find time after which the potential difference across it will be V_(0)? |
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Answer» Solution :When the slab is REMOVED, the potential difference across CAPACITOR increases to `kV_(0)` `CV_(0)=kCV_(0)e^(-(t)/(tau))` as `q_(0)=KCV_(0)` `(1)/(k)=e^(-(t)/(tau)) IMPLIES k=e^((t)/(tau))` `:. ln k=(1)/(tau) implies t= tau LNK` |
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