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(D) Let $C$ be the initial capacitance. Action $XW$: Connecting to battery $(V=E)$ and inserting dielectric $(K)$ results in charge $Q = KCE$ and elec

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(D) Let $C$ be the initial capacitance. Action $XW$: Connecting to battery $(V=E)$ and inserting dielectric $(K)$ results in charge $Q = KCE$ and electric field $E_{field} = E/d$. Action $WX$: Inserting dielectric $(K)$ and then connecting to battery $(V=E)$ results in charge $Q = KCE$ and electric field $E_{field} = E/d$. Thus,$XW$ and $WX$ result in the same electric field. Action $XWY$: Connecting to battery $(Q=CE)$,inserting dielectric $(Q=KCE)$,then disconnecting battery ($Q$ remains $KCE$). Action $XYW$: Connecting to battery $(Q=CE)$,disconnecting battery $(Q=CE)$,then inserting dielectric ($Q$ remains $CE$). Since $K > 1$,charge $KCE > CE$,so $XWY$ has more charge. Energy $U = Q^2 / (2C_{final})$. For $WXY$: $U = (KCE)^2 / (2KC) = (KCE^2)/2$. For $XYW$: $U = (CE)^2 / (2KC) = (CE^2)/(2K)$. Since $K > 1$,$(KCE^2)/2 > (CE^2)/(2K)$,so $WXY$ has more energy.

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