Question 1
A capacitor is made with a polymeric dielectric having an \(\varepsilon_r\) of 2.26 and a dielectric breakdown strength of \(50 \text{ kV/cm}\). The permittivity of free space is \(8.85 \text{ pF/m}\). If the rectangular plates of the capacitor have a width of \(20 \text{ cm}\) and a length of \(40 \text{ cm}\), then the maximum electric charge in the capacitor is:
Solution
The maximum charge \(q\) is given by the product of capacitance \(C\) and the breakdown voltage \(V\):
Equation
\[q = CV\]
We know that capacitance \(C = \frac{\varepsilon A}{d}\), so:
Equation
\[q = \left( \frac{\varepsilon A}{d} \right) \times V = \varepsilon A \left( \frac{V}{d} \right)\]
Since the electric field strength \(E = \frac{V}{d}\), we can substitute this directly:
Equation
\[q = \varepsilon A E = (\varepsilon_r \varepsilon_0) A E\]
Given values:
- Relative permittivity, \(\varepsilon_r = 2.26\)
- Permittivity of free space, \(\varepsilon_0 = 8.85 \text{ pF/m} = 8.85 \times 10^{-12} \text{ F/m} = 8.85 \times 10^{-14} \text{ F/cm}\)
- Electric field strength, \(E = 50 \text{ kV/cm} = 50 \times 10^3 \text{ V/cm}\)
- Area, \(A = \text{width} \times \text{length} = 20 \text{ cm} \times 40 \text{ cm} = 800 \text{ cm}^2\)
Calculation:
Equation
\[\begin{aligned}
q &= (2.26) \times (8.85 \times 10^{-14} \text{ F/cm}) \times (800 \text{ cm}^2) \times (50 \times 10^3 \text{ V/cm}) \\
q &= 2.26 \times 8.85 \times 10^{-14} \times 800 \times 50000 \\
q &= 2.26 \times 8.85 \times 10^{-14} \times (4 \times 10^7) \\
q &= 2.26 \times 8.85 \times 4 \times 10^{-7} \\
q &= 79.99 \times 10^{-7} \text{ C} \\
q &\approx 80 \times 10^{-7} \text{ C} \\
q &= 8 \times 10^{-6} \text{ C} \\
q &= 8 \mu\text{C}
\end{aligned}\]
Final Answer
Correct answer: C.