Match the following :
Diagonally checked rectangle represents dielectric slab.

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Solution

For A: as battery is connected to the capacitor so potential $V$ remains constant. When dielectric slab is removed from the capacitor, capacitance $C^{'} = k C$ becomes $C$ and energy $U = 1 / 2 k C V^{2}$ becomes $1 / 2 C V^{2}$ . Thus capacitance and store energy decreases. As charge $Q = C V$ so charge on the plates decreases. so, $A \to 4$
For B: here also potential $V$ remains constant. When dielectric slab is inserted into the capacitor, capacitance $C$ becomes $k C$ and energy $U = 1 / 2 C V^{2}$ becomes $1 / 2 k C V^{2}$ . Thus capacitance and store energy increases. As charge $Q = C V$ so charge on the plates increases. so $B \to 2, 3$
For C: as no cell is connected to capacitor so charge remains constant. When dielectric slab is removed from the capacitor, capacitance $C^{'} = k C$ becomes $C$ and energy $U = 1 / 2 (Q^{2} / k C)$ becomes $1 / 2 (Q^{2} / C)$ . Thus capacitance decreases and store energy increases. As potential $V = Q / C$ so potential between the plates increases. so, $C \to 1, 3$
For D: here also Q remains constant. When dielectric slab is insert into the capacitor, capacitance $C$ becomes $k C$ and energy $U = 1 / 2 (Q^{2} / C)$ becomes $1 / 2 (Q^{2} / k C)$ . Thus capacitance increases and store energy decreases. As potential $V = Q / C$ so potential between the plates decreases. so, $D \to 2$

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Match the following :Diagonally checked rectangle represents dielectric slab.

Match the following :
Diagonally checked rectangle represents dielectric slab.