A 60   μF parallel plate capacitor whose plates are separated by 6   mm is charged to 250…

Capacitance of capacitor after inserting a slab of dielectric constant $5$and of thickness $t=3 \mathrm{mm}$ is, $C\text{'}=C\left[\frac{1}{1-{\displaystyle \frac{t}{d}}+{\displaystyle \frac{t}{kd}}}\right]$

$$\begin{gathered} \Rightarrow C\text{'}=12 \mathrm{\mu F}\left[\frac{1}{1-{\displaystyle \frac{3 \mathrm{mm}}{6 \mathrm{mm}}}+{\displaystyle \frac{3 \mathrm{mm}}{5\times 6 \mathrm{mm}}}}\right]=12 \mathrm{\mu F}\times \frac{20}{12} \\ \Rightarrow \mathrm{C}\text{'}=20 \mathrm{\mu F} \end{gathered}$$

Now, after removing the charging source from the capacitor the charge remain same, $q=CV=C\text{'}V\text{'}$

$$\begin{gathered} \Rightarrow CV=C\text{'}V\text{'} \\ \Rightarrow 12 \mathrm{\mu F}\times 250 \mathrm{V}=20 \mathrm{\mu F}\times V\text{'} \\ \Rightarrow V\text{'}=150 \mathrm{V} \end{gathered}$$

Thus, the change in the potential difference across the capacitor is$\left(250 \mathrm{V}-150\mathrm{V}\right)=100 \mathrm{V}$