 Solution

The quoted capacitance is that of the filled parallel-plate capacitor, $C_i=1.0\times 10^{-6}\mathrm{F}$. Disconnecting the source imposes charge conservation on the isolated plates:On removing the dielectric, the capacitance falls by its relative permittivity, so $C_f=C_i/5=2.0\times 10^{-7}\mathrm{F}$. The capacitor energy at fixed electric charge is $U=Q^2/(2C)$. ThereforeandThe increase is supplied by the external work needed to withdraw the dielectric against its attraction into the electric field, in the lossless quasistatic limit. The final electric potential difference is $Q/C_f=12500\mathrm{V}$, not the original charging value.