The dielectric slab is introduced between the plates of a parallel plate charged capacitor. Which one of the following quantities will not change?

Only the amount of charge stored in the capacitor will remain constant when the dielectric is inserted between the plates (isolated system).
Charge will redistribute, such on the plate \(Q\left(1-\frac{1}{K}\right)\) and \(\frac{Q}{K}\) on the dielectric interface.
The electric field present between the plates of capacitor changes because it varies with the dielectric constant:
\(E^{\prime}=\frac{Q}{4 \pi \varepsilon_0 K}=\frac{E}{K}\)
The capacitance increases as:
\(C^{\prime}=\frac{K \varepsilon_0 A}{d}=K C\)
The dielectric constant also affects the potential difference as:
\(\begin{aligned} & Q=C^{\prime} V^{\prime}=C K V^{\prime}=C V \\ & \therefore V^{\prime}=\frac{V}{K}\end{aligned}\)
and the energy stored in the capacitor:
\(U^{\prime}=\frac{1}{2} \frac{Q^2}{C^{\prime}}=\frac{U}{K}\)