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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
charge $CE(k - 1)$ flows through the cell
energy $E^2C(k - 1)$ is absorbed by the cell.
the external agent has to do $\frac{1}{2}E^2 C(k - 1)$ amount of work to take the slab out.
all of the above
Solution
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 $=\frac{1}{2}(\mathrm{K}-1) \mathrm{CV}^{2}$
