Question

A parallel plate capacitor of plate separation D is charged to a potential difference V . A Dielectric slab of thickness D and dielectric constant K is introduced between the plates while the battery remains connected to the plates.
1 Find the ratio of energy stored in the capacitor after and before the dielectric is introduced. give the physical explanation for this change in stored energy?
2 what happens to the charge in the capacitor?

Answer 1

Dear Student,


$$\begin{gathered} chargebeforethedielectricintroduction \\ Q=CV,E=\frac{C{V}^2}{2} \\ whenthedielectricisinroduced \\ Q\text{'}=KCV,E\text{'}=\frac{KC{V}^2}{2} \\ \frac{E\text{'}}{E}=\frac{\frac{KC{V}^2}{2}}{\frac{C{V}^2}{2}}=K \\ hencetheenergystoredincreased. \\ thechargeinthecapacitorincreasedtoKCV. \\ Regards \end{gathered}$$