Question

A parallel-plate capacitor of area $A$, plate separation d and capacitance $C$ is filled with four dielectric materials having dielectric constants $k_1, k_2, k_3 $ and $ k_4$ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $k$ is given by

• A. $k = k _1 + k_2 + k_3 + 3k_4$
• B. $k = \frac{2}{3} (k_1 + k_2 + k_3) + 2k_4$
• C. $\frac{2}{k} = \frac{3}{k_1 + k_2 + k_3} + \frac{1}{k_4}$
• D. $\frac{1}{k} = \frac{1}{k_1 } + \frac{1}{k_2} + \frac{1}{k_3} + \frac{3}{2 k_4}$

Answer 1

The Correct Option is C

Capacitance and Spring Constants Calculation 

We are given the following equations and need to solve for the relationship between different parameters:

1. Capacitance Formula

The formula for the total capacitance of two capacitors in parallel is given by:

\(\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}\)

2. Rearranged Equation for Capacitor in Terms of Spring Constants

The equation involving spring constants and capacitance is:

\(\frac{d}{kA \varepsilon_0} = \frac{3d}{2(k_1 + k_2 + k_3)A \varepsilon_0} + \frac{d}{2k_4 A \varepsilon_0}\)

3. Simplifying the Equation

Next, we simplify the equation by factoring out \( \frac{d}{A \varepsilon_0} \):

\(\frac{d}{kA \varepsilon_0} = \frac{d}{A \varepsilon_0} \left[ \frac{3}{2(k_1 + k_2 + k_3)} + \frac{1}{2k_4} \right]\)

4. Final Simplified Expression

The final simplified form of the equation is:

\(\frac{2}{k} = \frac{3}{k_1 + k_2 + k_3} = \frac{1}{k_4}\)

Conclusion:

The equation shows the relationship between the spring constants \( k_1, k_2, k_3, k_4 \) and the overall constant \( k \). By solving for these constants, we can find the total system behavior in terms of its capacitance and spring constants.