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Step-by-Step Solution
Step 1: Identify the Two Capacitors in Series
Since the plate separation is 10 m, with 5 m of it filled by a dielectric (of constant K=10 ) and the remaining 5 m effectively air (or vacuum, K=1 ), we can view the setup as two capacitors connected in series:
1. Capacitor C_1 : Thickness of dielectric = 5 \,\text{m} , Dielectric constant = 10 .
2. Capacitor C_2 : Thickness of air = 5 \,\text{m} , Dielectric constant = 1 .
Step 2: Write the Formula for Each Capacitor
The capacitance of a parallel plate capacitor of thickness d , area A , and dielectric constant K is
C = \frac{K \,\varepsilon_0\, A}{d}.
Here, \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} , and A = 100 \text{ m}^2 .
Step 3: Calculate C_1 (Dielectric-Slab Portion)
For the dielectric portion ( d=5 \,\text{m} , K=10 ):
C_1
= \frac{10\,\varepsilon_0\, A}{5}
= \frac{10 \times 8.85\times 10^{-12} \times 100}{5}.
Simplifying,
10 \times 8.85 \times 10^{-12}
= 8.85 \times 10^{-11},
then
8.85 \times 10^{-11} \times 100
= 8.85 \times 10^{-9},
and dividing by 5,
C_1 = 1.77 \times 10^{-9}\,\text{F}
= 1770 \,\text{pF}.
Step 4: Calculate C_2 (Air-Gap Portion)
For the air portion ( d=5 \,\text{m} , K=1 ):
C_2
= \frac{\varepsilon_0\, A}{5}
= \frac{8.85\times 10^{-12} \times 100}{5}.
Similarly calculating,
8.85 \times 10^{-12} \times 100
= 8.85 \times 10^{-10},
and dividing by 5,
C_2 = 1.77 \times 10^{-10}\,\text{F}
= 177 \,\text{pF}.
Step 5: Combine the Two Capacitors in Series
For capacitors in series, the equivalent capacitance C_{\text{eq}} is given by
\frac{1}{C_\text{eq}}
= \frac{1}{C_1} + \frac{1}{C_2}.
Hence,
\frac{1}{C_\text{eq}}
= \frac{1}{1770\,\text{pF}} + \frac{1}{177\,\text{pF}}.
Numerically,
\frac{1}{1770} \approx 5.65 \times 10^{-4},
\quad
\frac{1}{177} \approx 5.65 \times 10^{-3}.
Summing,
5.65 \times 10^{-4} + 5.65 \times 10^{-3}
\approx 6.215 \times 10^{-3}.
Therefore,
C_{\text{eq}}
\approx \frac{1}{6.215 \times 10^{-3}}
= 1.61 \times 10^{2}\,\text{pF}
= 161\,\text{pF}.
Step 6: State the Final Answer
The resultant capacitance of the system to the nearest integer is
\boxed{161 \,\text{pF}.}
