Duration: 5:19

0 views

0

The capacitance of a parallel-plate capacitor is \( C_{0} \) when the region between the plates has air. This region is now filled with a dielectric slab of dielectric constant \( \kappa \). The capacitor is connected to a cell of emf \( \varepsilon \) and the slab is taken out.
(A) Charge \( \varepsilon \mathrm{C}_{0}(\kappa-1) \) flows through the cell
(B) Energy \( \varepsilon^{2} C_{0}(\kappa-1) \) is absorbed by the cell
(C) The energy stored in the capacitor is reduced by \( \varepsilon^{2} C_{0}(\kappa-1) \)
(D) The external agent has to do \( (1 / 2) \varepsilon^{2} C_{0}(\kappa-1) \) amount of work to take the slab out
📲PW App Link - https://bit.ly/YTAI_PWAP
🌐PW Website - https://www.pw.live