Physics - Capacitor Question with Solution | TestHub

Key Idea : After filling with dielectrics the two capacitors will be in parallel order. As shown, the two capacitors are connected in parallel. Initially the capacitance of capacitor$C=\frac{{\epsilon }_{0}A}{d}$ where $A$ is area of each plate and $d$ is the separation between the plates. After filling with dielectrics, we have two capacitors of capacitance. $C_1=\frac{K_1{\epsilon }_0(A/2)}{d}=\frac{8}{2}\frac{{\epsilon }_{0}A}{d}=\frac{4{\epsilon }_{0}A}{d}=4C$ and $C_2=\frac{K_2{\epsilon }_0(A/2)}{d}=\frac{4}{2}\frac{{\epsilon }_{0}A}{d}=\frac{2{\epsilon }_{0}A}{d}=2C$ Hence, their equivalent capacitance $\begin{array}{rl}{C}_{\mathrm{eq}}& =C_1+C_2\\ & =4C+2C=6C\end{array}$ ie, new capacitance will be six times of the original.