Physics - Capacitor Question with Solution | TestHub

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: $E^{\prime }=\frac{Q}{4\pi {\epsilon }_{0}K}=\frac{E}{K}$ The capacitance increases as: $C^{\prime }=\frac{K{\epsilon }_{0}A}{d}=KC$ The dielectric constant also affects the potential difference as: $\begin{array}{rl}& Q=C^{\prime }{V}^{\prime }=CK{V}^{\prime }=CV\\ & \therefore V^{\prime }=\frac{V}{K}\end{array}$ and the energy stored in the capacitor: $U^{\prime }=\frac{1}{2}\frac{Q^2}{C^{\prime }}=\frac{U}{K}$