Question
Two identical air filled parallel plate capacitors are charged to the same potential in the manner shown by closing the switch $S.$ If now the switch $S$ is opened and the space between the plates is filled with a dielectric of relative permittivity $ε_{r​},$ then:

Answer

After switch $S$ is opened, as the Capacitor $A$ is connected across the battery, its potential difference is fixed at steady state $($i.e., when capacitor is fully charged$).$
Capacitor $B$ is isolated, so its charge gets fixed. But as we insert the dielectric, its capacitance changes, thus its potential difference also changes.

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