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Step-by-Step Solution
Step 1: Identify the Original Capacitance
We are given a parallel plate capacitor with plate area $A$, plate separation $d = 2\,\text{m}$, and initial capacitance $C = 4\,\mu\text{F}$. From the parallel plate capacitor formula:
$C = \dfrac{\varepsilon_0 A}{d} \quad\Longrightarrow\quad \dfrac{\varepsilon_0 A}{d} = 4\,\mu\text{F}.$
Step 2: Understand the Effect of Filling Half the Thickness with Dielectric
When half of the space (by thickness) between the plates is filled with a dielectric of constant $K = 3$, we can think of the capacitor as two capacitors in series:
The first capacitor ($C_1$) has thickness $d/2$ with dielectric constant $K = 3$.
The second capacitor ($C_2$) has thickness $d/2$ with air (dielectric constant = 1).
Hence, $C_1$ and $C_2$ are connected in series.
Step 3: Calculate Capacitances $C_1$ and $C_2$
The general formula for a parallel plate capacitor of thickness $t$ and dielectric constant $K$ is
$C = \dfrac{K\,\varepsilon_0 A}{t}.$
For $C_1$, thickness $= d/2$ and dielectric constant $= 3$, so
$$
C_1 = \dfrac{3 \,\varepsilon_0 A}{(d/2)}
= \dfrac{2 \cdot 3 \,\varepsilon_0 A}{d}
= \dfrac{6\,\varepsilon_0 A}{d}.
$$
For $C_2$, thickness $= d/2$ and dielectric constant $= 1$, so
$$
C_2 = \dfrac{\varepsilon_0 A}{(d/2)}
= \dfrac{2\,\varepsilon_0 A}{d}.
$$
Step 4: Combine $C_1$ and $C_2$ in Series
For two capacitors in series, the equivalent capacitance $C_{\text{eq}}$ is given by
$$
\dfrac{1}{C_{\text{eq}}}
= \dfrac{1}{C_1} + \dfrac{1}{C_2}.
$$
Substituting $C_1$ and $C_2$:
$$
\dfrac{1}{C_{\text{eq}}}
= \dfrac{1}{\tfrac{6\,\varepsilon_0 A}{d}}
+ \dfrac{1}{\tfrac{2\,\varepsilon_0 A}{d}}
= \dfrac{d}{6\,\varepsilon_0 A}
+ \dfrac{d}{2\,\varepsilon_0 A}
= \dfrac{d}{2\,\varepsilon_0 A} \Bigl(\dfrac{1}{3} + 1\Bigr)
= \dfrac{d}{2\,\varepsilon_0 A} \times \dfrac{4}{3}.
$$
Hence,
$$
C_{\text{eq}}
= \dfrac{1}{\dfrac{d}{2\,\varepsilon_0 A} \times \dfrac{4}{3}}
= \dfrac{2\,\varepsilon_0 A \,\times\, 3}{4\,d}
= \dfrac{3}{2} \dfrac{\varepsilon_0 A}{d}.
$$
Step 5: Substitute the Known Value of $\dfrac{\varepsilon_0 A}{d}$
From Step 1, we have $\dfrac{\varepsilon_0 A}{d} = 4\,\mu\text{F}$. Therefore,
$$
C_{\text{eq}}
= \dfrac{3}{2} \times 4\,\mu\text{F}
= 6\,\mu\text{F}.
$$
Final Answer
The new capacitance of the system is $6\,\mu\text{F}$.
