Question

A parallel plate capacitor of area $A$, plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constants $k_{1},\, k_{2}$ and $k_{3}$ as shown. 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. $\frac{1}{k}=\frac{1}{k_{1}}+\frac{1}{k_{2}}+\frac{1}{2 k_{3}}$
• B. $\frac{1}{k}=\frac{1}{k_{1}+k_{2}}+\frac{1}{2 k_{3}}$
• C. $k=\frac{k_{1} k_{2}}{k_{1}+k_{2}}+2 k_{3}$
• D. $k=k_{1}+k_{2}+2 k_{3}$

Answer 1

Answer: (B)

$C_{1}=\frac{k_{1} \varepsilon_{0} \frac{A}{2}}{\left(\frac{d}{2}\right)}=\frac{k_{1} \varepsilon_{0} A}{d}$
$C_{2}=\frac{k_{2} \varepsilon_{0} \frac{A}{2}}{\left(\frac{d}{2}\right)}=\frac{k_{2} \varepsilon_{0} A}{d}$
and $C_{3}=\frac{k_{3} \varepsilon_{0} A}{\left(\frac{d}{2}\right)}=\frac{2 k_{3} \varepsilon_{0} A}{d}$
$\frac{1}{C_{e q}}=\frac{1}{C_{1}+C_{2}}+\frac{1}{C_{3}}$
$=\frac{1}{\frac{\varepsilon_{0} A}{d}\left(k_{1}+k_{2}\right)}+\frac{1}{\frac{\varepsilon_{0} A}{d} \times 2 k_{3}}$
$\frac{1}{C_{e q}}=\frac{d}{\varepsilon_{0} A}\left[\frac{1}{k_{1}+k_{2}}+\frac{1}{2 k_{3}}\right]$
$C_{e q}=\left[\frac{1}{k_{1}+k_{2}}+\frac{1}{2 k_{3}}\right]^{-1} \cdot \frac{\varepsilon_{0} A}{d}$
So $k_{e q}=\left[\frac{1}{k_{1}+k_{2}}+\frac{1}{2 k_{3}}\right]^{-1}$