The capacitance of a parallel-plate capacitor is C0 when the region between the plates has air. This region is now filled with a dielectric slab of dielectric constant K. The capacitor is connected to a cell emf ε1 and the slab is taken out.
A
Charge εC0(K−1) flows through the cell.
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B
Energy ε2C0(K−1) is absorbed by the cell.
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C
The energy stored in the capacitor is reduced by ε2C0(K−1).
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D
The external agent has to do 12ε2C0(K−1) amount of work to take the slab out.
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Solution
The correct options are A Charge εC0(K−1) flows through the cell. B Energy ε2C0(K−1) is absorbed by the cell. D The external agent has to do 12ε2C0(K−1) amount of work to take the slab out. The capacitance of capacitor with dielectric is C=KC0 Initial charge on capacitor is q=Cε=KC0ε As cell is still connect so potential remain constant. after slab is taken out the charge on capacitor is q′=C0ε The charge flows through the cell is Δq=q−q′=KC0ε−C0ε=C0ε(K−1) Energy absorbed by cell E=Δqε=C0ε2(K−1) Energy stored in capacitor U=12qε−12q′ε=12C0ε2(K−1) Work done W=E−U=C0ε2(K−1)−12C0ε2(K−1)=12C0ε2(K−1)