Step 1: Conceptualization:
This problem concerns a parallel plate capacitor loaded with a dielectric. The objective is to determine the dielectric constant (k) based on provided physical parameters. Capacitance can be quantified by the charge (Q) and voltage (V), as well as by the capacitor's physical attributes and the dielectric material.
Step 2: Foundational Equations:
Capacitance (\(C\)) is defined as the ratio of charge (\(Q\)) to potential difference (\(V\)):
\[ C = \frac{Q}{V} \]For a parallel plate capacitor containing a dielectric, capacitance is also expressed as:
\[ C = \frac{k \varepsilon_0 A}{d} \]Here, \(k\) denotes the dielectric constant, \(\varepsilon_0\) represents the permittivity of free space (\(8.854 \times 10^{-12} \, \text{F/m}\)), \(A\) is the area of the plates, and \(d\) is the separation distance between them. Equating these two expressions for \(C\) will yield \(k\).
Step 3: Procedural Breakdown:
Input Data:
Plate area, \(A = 200 \, \text{cm}^2 = 200 \times 10^{-4} \, \text{m}^2\).
Plate separation, \(d = 2.0 \, \text{mm} = 2.0 \times 10^{-3} \, \text{m}\).
Charge, \(Q = 0.06 \, \mu\text{C} = 0.06 \times 10^{-6} \, \text{C}\).
Potential difference, \(V = 60 \, \text{V}\).
Execution:
Calculate capacitance from charge and voltage:
\[ C = \frac{Q}{V} = \frac{0.06 \times 10^{-6} \, \text{C}}{60 \, \text{V}} = 1 \times 10^{-9} \, \text{F} \]Rearrange the parallel plate capacitor formula to isolate the dielectric constant \(k\):
\[ k = \frac{C d}{\varepsilon_0 A} \]Substitute the provided values:
\[ k = \frac{(1 \times 10^{-9} \, \text{F}) \times (2.0 \times 10^{-3} \, \text{m})}{(8.854 \times 10^{-12} \, \text{F/m}) \times (200 \times 10^{-4} \, \text{m}^2)} \]\[ k = \frac{2 \times 10^{-12}}{8.854 \times 10^{-12} \times 2 \times 10^{-2}} \]\[ k = \frac{2 \times 10^{-12}}{17.708 \times 10^{-14}} = \frac{2}{0.17708} \approx 11.294 \]
Step 4: Conclusion:
The dielectric constant is computed to be approximately 11.3.