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Step-by-Step Solution
Step 1: Identify the Geometry of the Problem
We are given a parallel plate capacitor of plate area 40 cm$^2$, with a plate separation of 2 mm. A dielectric of thickness 1 mm and dielectric constant $K = 5$ occupies part of the gap, and the remaining 1 mm is filled with air. Essentially, we have two separate capacitors in series:
Capacitor 1: Dielectric medium of thickness $t = 1 \times 10^{-3}\,\text{m}$ and dielectric constant $K = 5$.
Capacitor 2: Air gap of thickness $(d - t) = 1 \times 10^{-3}\,\text{m}$ (since the total gap $d = 2 \times 10^{-3}\,\text{m}$).
Step 2: Express Individual Capacitances
Let $A$ be the plate area in m$^2$ and $\epsilon_0$ be the permittivity of free space. First, we convert the area to SI units:
$40 \,\text{cm}^2 = 40 \times 10^{-4}\,\text{m}^2$.
The capacitances of the two sections are:
$C_1 = \dfrac{K \, \epsilon_0 \, A}{t}, \quad C_2 = \dfrac{\epsilon_0 \, A}{(d - t)}.$
Step 3: Use the Formula for Series Combination
Since these two capacitors are in series, the equivalent capacitance $C_{\mathrm{eq}}$ is given by:
$\dfrac{1}{C_{\mathrm{eq}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}.$
Step 4: Substitute the Expressions of $C_1$ and $C_2$
Substitute $C_1 = \dfrac{K \,\epsilon_0 \,A}{t}$ and $C_2 = \dfrac{\epsilon_0 \,A}{(d - t)}$ into the series formula:
$\dfrac{1}{C_{\mathrm{eq}}}
= \dfrac{1}{\dfrac{K \, \epsilon_0 \, A}{t}} + \dfrac{1}{\dfrac{\epsilon_0 \, A}{(d - t)}}
= \dfrac{t}{K \, \epsilon_0 \, A} + \dfrac{(d - t)}{\epsilon_0 \, A}.$
Step 5: Plug in the Numerical Values
Here, $t = 1 \times 10^{-3}\,\text{m}$, $d - t = 1 \times 10^{-3}\,\text{m}$, $A = 40 \times 10^{-4}\,\text{m}^2$, and $K = 5$. Substituting:
$\dfrac{1}{C_{\mathrm{eq}}}
= \dfrac{1 \times 10^{-3}}{5 \,\epsilon_0 \,(40 \times 10^{-4})}
+ \dfrac{1 \times 10^{-3}}{\epsilon_0 \,(40 \times 10^{-4})}.$
Upon simplification, this becomes:
$\dfrac{1}{C_{\mathrm{eq}}}
= \dfrac{1}{20\,\epsilon_0}
+ \dfrac{1}{4\,\epsilon_0}
= \dfrac{1 + 5}{20\,\epsilon_0}
= \dfrac{6}{20\,\epsilon_0}
= \dfrac{3}{10\,\epsilon_0}.$
Hence,
$C_{\mathrm{eq}}
= \dfrac{10}{3}\,\epsilon_0 \,\text{F}.$
Step 6: Final Answer
Therefore, the equivalent capacitance of the system is
$\displaystyle \dfrac{10}{3}\,\epsilon_0\,\text{F}.$
Supporting Diagram
