The capacitance of a parallel plate capacitor is C when the region between the plate has air. This region is now filled with a dielectric

Q: The capacitance of a parallel plate capacitor is $C$ when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant $k$. The capacitor is connected to a cell of $emf$ $E$, and the slab is taken out

d Charge $=\mathrm{KCV}-\mathrm{CV}=(\mathrm{K}-1) \mathrm{CV}$ Energy absorbed $=(\mathrm{K}-1) \mathrm{CV}^{2}$ Energy $=\frac{1}{2} \times \mathrm{K} \times \mathrm{C} \times \mathrm{V}^{2}=\frac{(\mathrm{K}-1)}{2} \cdot \mathrm{CV}^{2}$ Now $\frac{1}{2} \times \mathrm{K} \times \mathrm{C} \times \mathrm{V}^{2}+$ work $=\frac{1}{2} \times \mathrm{C} \times \mathrm{V}^{2}=(\mathrm{K}-1) \mathrm{CV}^{2}$ $\Rightarrow$ Work $=\frac{1}{2}(\mathrm{K}-1) \mathrm{CV}^{2}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

The capacitance of a parallel plate capacitor is $C$ when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant $k$. The capacitor is connected to a cell of $emf$ $E$, and the slab is taken out

Acharge $CE(k - 1)$ flows through the cell
Benergy $E^2C(k - 1)$ is absorbed by the cell.
Cthe external agent has to do $\frac{1}{2}E^2 C(k - 1)$ amount of work to take the slab out.
D
all of the above

Advanced

STD 11 Science
NEET
STD 12 - 2. Electric potential and capacitance
Physics

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