Question 1
A capacitor consists of two metal plates each \(500 \times 500 \text{ mm}^2\) and spaced \(6 \text{ mm}\) apart. The space between the metal plates is filled with a glass plate of \(4 \text{ mm}\) thickness and a layer of paper of \(2 \text{ mm}\) thickness. The relative permittivity of the glass and paper are 8 and 2 respectively. Neglecting the fringing effect, the capacitance will be (Given that \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}\)).
Solution
Given:
- Area (\(A\)): \(500 \times 500 \text{ mm}^2 = 250,000 \text{ mm}^2 = 0.25 \text{ m}^2\)
- Permittivity (\(\varepsilon_0\)): \(8.85 \times 10^{-12} \text{ F/m}\)
- Dielectric Layers:
- Glass: \(d_g = 4 \text{ mm} = 4 \times 10^{-3} \text{ m}, \quad \varepsilon_{r,g} = 8\)
- Paper: \(d_p = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}, \quad \varepsilon_{r,p} = 2\)
The arrangement acts as two capacitors in series. The equivalent capacitance \(C\) for composite dielectrics in series is given by:
Equation
\[C = \frac{\varepsilon_0 A}{\frac{d_g}{\varepsilon_{r,g}} + \frac{d_p}{\varepsilon_{r,p}}}\]
Substituting the values:
Equation
\[\begin{aligned}
C &= \frac{(8.85 \times 10^{-12}) \times (0.25)}{\frac{4 \times 10^{-3}}{8} + \frac{2 \times 10^{-3}}{2}} \\
C &= \frac{2.2125 \times 10^{-12}}{(0.5 \times 10^{-3}) + (1.0 \times 10^{-3})} \\
C &= \frac{2.2125 \times 10^{-12}}{1.5 \times 10^{-3}} \\
C &= 1.475 \times 10^{-9} \text{ F} \\
C &= 1475 \text{ pF}
\end{aligned}\]
Final Answer
Correct answer: B.