Potential drop across a dielectric capacitor with battery

Question

I have a capacitor (C) connected to a battery of emf 100V. Hence potential drop across my capacitor is 100V. Now I introduce a dielectric between the plates of capacitor which increases the capacitance to C'. My book says that placing a dielectric inside the capacitor reduces the potential difference between the plates by reducing Electric field strength between them. But when finding the increase in energy due to dielectric, they used
(1/2)(C'-C)(100V)²

Why do we use 100V as potential drop, eventhough we know that dielectric reduced the potential difference between the plates?

Bob D · Answer

Your book is talking about a capacitor that is changed to 100 volts and not connected to a battery. Inserting a dielectric increases the capacitance $C$. Since the charge $Q$ doesn’t change (it has no where to go) the voltage $V$ across the capacitor drops due to the relationship
$$C=frac{Q}{V}$$
if you increase $C$ and $Q$ is constant $V$ has to be less.
The energy equation applies if you insert the dielectric while capacitor is connected to the battery. The voltage is fixed at the emf of the battery. Increasing the capacitance with the voltage fixed increases the charge on the capacitor per the above equation. The battery is what increases the charge.
Hope this helps
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Potential drop across a dielectric capacitor with battery

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