Question

Find out the formula for the capacitance of the parallel plate capacitor shown in the figure. Area of the plates is $A$ and thicknesses of the dielectric slabs between the plates are $d_1$ and $d_2$, and their dielectric constants are $K_1$ and $K_2$ respectively.

Hint:

For multiple dielectrics in series, treat them as series capacitors. The effective plate separation is replaced by $\dfrac{d_1}{K_1} + \dfrac{d_2}{K_2}$.

Answer 1

Step 1: Recall formula for capacitance.
For a parallel plate capacitor with dielectric of thickness $d$ and constant $K$: \[ C = \frac{\epsilon_0 K A}{d}. \]
Step 2: Observation of system.
Here, the capacitor contains two dielectric slabs of thicknesses $d_1, d_2$ and constants $K_1, K_2$, placed in series (stacked one after another between plates).
Step 3: Equivalent capacitance for series connection.
For series capacitors: \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}. \]
Step 4: Individual capacitances.
\[ C_1 = \frac{\epsilon_0 K_1 A}{d_1}, \quad C_2 = \frac{\epsilon_0 K_2 A}{d_2}. \]
Step 5: Substitute in series formula.
\[ \frac{1}{C} = \frac{d_1}{\epsilon_0 K_1 A} + \frac{d_2}{\epsilon_0 K_2 A}. \]
Step 6: Simplify.
\[ \frac{1}{C} = \frac{1}{\epsilon_0 A}\left(\frac{d_1}{K_1} + \frac{d_2}{K_2}\right). \] \[ C = \frac{\epsilon_0 A}{\dfrac{d_1}{K_1} + \dfrac{d_2}{K_2}}. \]
Step 7: Conclusion.
Thus, the capacitance of the system is: \[ C = \frac{\epsilon_0 A}{\dfrac{d_1}{K_1} + \dfrac{d_2}{K_2}}. \]