The capacitance of a parallel plate capacitor is C0 when the plates have air between them. This region is now filled with a dielectric slab of dielectric constant K and capacitor is connected with a battery of emf E and zero internal resistance. Now slab is taken out, then during removal of slab:
1.) Charge EC0(K−1) flows through the cell.
2.) Energy E2C0(K−1) is absorbed by cell.
3.) The energy stored in the capacitor is reduced by E2C0(K−1).
4.) The external agent has to do E2C0(K−1) amount of work to take out the slab.
Select the correct phenomenon happening during the process:
Why this question? Caution: If a positive charge flows through a cell such that it leaves the +ve terminal of battery and enters through its -ve terminal, then work is done by battery i.e. Wbattery>0 and for the reverse case Wbattery<0. |