Question:
A parallel plate capacitor has a separation between plates of 0.885 mm. It has a capacitance of $1\,\mu\text{F}$ when the space between the plates is filled with an insulating material of resistivity $1 \times 10^{13}\,\Omega\,\text{m}$ and resistance $17.7 \times 10^{14}\,\Omega$. The relative permittivity of the insulating material is $\alpha \times 10^7$. The value of $\alpha$ is \underline{\hspace{2cm}}.
A parallel plate capacitor has a separation between plates of 0.885 mm. It has a capacitance of $1\,\mu\text{F}$ when the space between the plates is filled with an insulating material of resistivity $1 \times 10^{13}\,\Omega\,\text{m}$ and resistance $17.7 \times 10^{14}\,\Omega$. The relative permittivity of the insulating material is $\alpha \times 10^7$. The value of $\alpha$ is \underline{\hspace{2cm}}.
Updated On: Apr 13, 2026
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Correct Answer: 2
Solution and Explanation
Step 1: Understanding the Concept:
For a medium of given resistivity (\(\rho\)) and permittivity (\(\epsilon\)), the product of its resistance (\(R\)) and capacitance (\(C\)) is a constant that depends only on the material properties, not the geometry. This is derived from the relations \(R = \rho \frac{d}{A}\) and \(C = \frac{\epsilon A}{d}\).
Step 2: Key Formula or Approach:
The fundamental relationship is: \[ RC = \rho \epsilon = \rho \epsilon_0 \epsilon_r \]
Step 3: Detailed Explanation:
Given values: \(R = 17.7 \times 10^{14}\) \(\Omega\), \(C = 1 \times 10^{-6}\) F, \(\rho = 1 \times 10^{13}\) \(\Omega \cdot \text{m}\), \(\epsilon_0 = 8.85 \times 10^{-12}\) F/m. Substituting these into the formula: \[ (17.7 \times 10^{14})(1 \times 10^{-6}) = (1 \times 10^{13})(8.85 \times 10^{-12}) \epsilon_r \] \[ 17.7 \times 10^8 = (8.85 \times 10^1) \epsilon_r \] \[ \epsilon_r = \frac{17.7 \times 10^8}{88.5} \] \[ \epsilon_r = 0.2 \times 10^8 = 2 \times 10^7 \] Comparing with \(\alpha \times 10^7\), we get \(\alpha = 2\).
Step 4: Final Answer:
The value of \(\alpha\) is 2.
For a medium of given resistivity (\(\rho\)) and permittivity (\(\epsilon\)), the product of its resistance (\(R\)) and capacitance (\(C\)) is a constant that depends only on the material properties, not the geometry. This is derived from the relations \(R = \rho \frac{d}{A}\) and \(C = \frac{\epsilon A}{d}\).
Step 2: Key Formula or Approach:
The fundamental relationship is: \[ RC = \rho \epsilon = \rho \epsilon_0 \epsilon_r \]
Step 3: Detailed Explanation:
Given values: \(R = 17.7 \times 10^{14}\) \(\Omega\), \(C = 1 \times 10^{-6}\) F, \(\rho = 1 \times 10^{13}\) \(\Omega \cdot \text{m}\), \(\epsilon_0 = 8.85 \times 10^{-12}\) F/m. Substituting these into the formula: \[ (17.7 \times 10^{14})(1 \times 10^{-6}) = (1 \times 10^{13})(8.85 \times 10^{-12}) \epsilon_r \] \[ 17.7 \times 10^8 = (8.85 \times 10^1) \epsilon_r \] \[ \epsilon_r = \frac{17.7 \times 10^8}{88.5} \] \[ \epsilon_r = 0.2 \times 10^8 = 2 \times 10^7 \] Comparing with \(\alpha \times 10^7\), we get \(\alpha = 2\).
Step 4: Final Answer:
The value of \(\alpha\) is 2.
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