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Step-by-Step Solution
Step 1: Identify the Given Quantities
• Rated voltage of the capacitor, $V = 500 \text{ V}$
• Maximum electric field the dielectric can withstand, $E_\text{max} = 10^6 \text{ V/m}$
• Plate area, $A = 10^{-4} \text{ m}^2$
• Desired capacitance, $C = 15 \text{ pF} = 15 \times 10^{-12} \text{ F}$
• Permittivity of free space, $\varepsilon_0 = 8.86 \times 10^{-12}\, \text{C}^2 \text{N}^{-1} \text{m}^{-2}$
Step 2: Relate Distance Between Plates to Rated Voltage and Electric Field
The dielectric breakdown condition implies that the maximum electric field $E_\text{max}$ occurs when the capacitor is at its rated voltage $V$. Hence,
$$
d = \frac{V}{E_\text{max}} = \frac{500}{10^6} = 5 \times 10^{-4}\,\text{m}.
$$
Step 3: Write the Formula for the Capacitance of a Parallel Plate Capacitor
The capacitance of a parallel plate capacitor with a dielectric of constant $k$ is:
$$
C = \frac{k\,\varepsilon_0\,A}{d}.
$$
Therefore, rearranging to solve for $k$ gives:
$$
k = \frac{C\,d}{\varepsilon_0\,A}.
$$
Step 4: Substitute the Known Values
Substitute $C = 15 \times 10^{-12}\,\text{F}$, $d = 5 \times 10^{-4}\,\text{m}$,
$\varepsilon_0 = 8.86 \times 10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$, and $A = 10^{-4}\,\text{m}^2$:
$$
k
= \frac{(15 \times 10^{-12}) \times (5 \times 10^{-4})}{(8.86 \times 10^{-12}) \times (10^{-4})}
= \frac{15 \times 5}{8.86}
= 8.465 \approx 8.5.
$$
Step 5: Conclude the Dielectric Constant
Thus, the dielectric constant of the material is
$$
k \approx 8.5.
$$
