A capacitor is charged by using a battery,whi | vedclass

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Question

(C) When a capacitor is charged and the battery is disconnected,the charge $Q$ on the plates remains constant because there is no path for the charge

Vedclass · Accepted Answer

Decrease in the potential difference across the plates,reduction in stored energy,but no change in the charge on the plates.

Vedclass · Answer

(C) When a capacitor is charged and the battery is disconnected,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.When a dielectric slab of dielectric constant $K$ is inserted between the plates,the capacitance increases from $C$ to $C' = KC$.Since $Q = CV$,the new potential difference $V'$ is given by $V' = Q / C' = Q / (KC) = V / K$. Since $K > 1$,the potential difference decreases.The stored energy $U$ is given by $U = Q^2 / (2C)$. The new energy $U'$ is $U' = Q^2 / (2C') = Q^2 / (2KC) = U / K$. Since $K > 1$,the stored energy decreases.Therefore,the potential difference decreases,the stored energy decreases,and the charge remains constant.

List-$I$	List-$II$
$(P). PbS + 2PbO \xrightarrow{\Delta} 3Pb + SO_2 \uparrow$	$(1).$ Smelting
$(Q). PbS + \frac{3}{2}O_2 \xrightarrow{\Delta} PbO + SO_2 \uparrow$	$(2).$ Self reduction
$(R). PbCO_3 \xrightarrow{\Delta} PbO + CO_2 \uparrow$	$(3).$ Calcination
$(S). PbO + C \xrightarrow{\Delta} Pb + CO \uparrow$	$(4).$ Roasting

$A$ capacitor is charged by using a battery,which is then disconnected. $A$ dielectric slab is then inserted between the plates. What is the result?

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