Duration: 7:38

1 views

0

The capacitance of a parallel plate capacitor is \( \mathrm{C} \) when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant \( \mathrm{k} \). The capacitor is connected to a cell of emf \( \mathrm{E} \),
P and the slab is taken out
(A) charge \( \mathrm{CE}(\mathrm{k}-1) \) flows through the cell
W.
(B) energy \( \mathrm{E}^{2} \mathrm{C}(\mathrm{k}-1) \) is absorbed by the cell.
(C) the ener jy stored in the capacitor is reduced by \( \mathrm{E}^{2} \mathrm{C}(\mathrm{k}-1) \)
(D) the external agent has to do \( \frac{1}{2} \mathrm{E}^{2} \mathrm{C}(\mathrm{k}-1) \) amount of work to take the slab out.
📲PW App Link - https://bit.ly/YTAI_PWAP
🌐PW Website - https://www.pw.live