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Following operations can be performed on a capacitor :
X – connect the capacitor to a battery of emf , E.
Y – disconnect the battery
Z – reconnect the battery with polarity reversed.
W – insert a dielectric slab in the capacitor

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In XYZ (perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed.

The charge appearing on the capacitor is greater after the action XWY than after the action XYW.

The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.

The electric field in the capacitor after the action XW is the same as that after WX.

answer is B, C, D.

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Detailed Solution

$\begin{matrix} U_{x} = \frac{1}{2} {CE}^{2} \\ E_{x} = Ed \end{matrix}$

$U_{y} = \frac{1}{2} {CE}^{2}$

The Change in

$∆ Q_{X Y Z} = |Q + Q| = 2 Q$

$U_{Z} = \frac{1}{2} {CE}^{2}$

$\begin{matrix} C_{1} = KC \\ U_{1} = {KU}_{X} \\ E_{W} = E_{x} = Ed \end{matrix}$

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