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 figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constants 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. \(\frac{1}{k}=\frac{1}{k_{1}}+\frac{1}{k_{2}}+\frac{1}{k_{3}}+\frac{3}{2k_{4}}\)

Hint:

Treat the three slabs in the upper half as parallel capacitors and then combine the two halves in series.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The upper half, of thickness \(\frac{d}{2}\), has three slabs of area \(\frac{A}{3}\) each side by side, so they are in parallel. The lower half has one dielectric \(k_{4}\) over the full area.

Step 2: Calculation
\(C_{up}=\frac{2\epsilon_{0}A(k_{1}+k_{2}+k_{3})}{3d}\) and \(C_{low}=\frac{2\epsilon_{0}Ak_{4}}{d}\). In series: \(\frac{1}{C}=\frac{3d}{2\epsilon_{0}A(k_{1}+k_{2}+k_{3})}+\frac{d}{2\epsilon_{0}Ak_{4}}\).

Step 3: Compare the options
For one dielectric, \(\frac{1}{C}=\frac{d}{\epsilon_{0}AK}\). Equating and multiplying by \(2\) gives \(\frac{2}{K}=\frac{3}{k_{1}+k_{2}+k_{3}}+\frac{1}{k_{4}}\), the third option.

Final Answer:
The equivalent dielectric constant satisfies \(\frac{2}{k}=\frac{3}{k_{1}+k_{2}+k_{3}}+\frac{1}{k_{4}}\). \[ \boxed{\text{(C) }\dfrac{2}{k}=\dfrac{3}{k_{1}+k_{2}+k_{3}}+\dfrac{1}{k_{4}}} \]