Question

A parallel plate capacitor having plate area 200 cm² and separation 2.0 mm holds a charge of 0.06 µC on applying a potential difference of 60 V. The dielectric constant of the material filled in between the plates is.

• A. 0.113
• B. 1.13
• C. 11.3
• D. 113

Hint:

Always ensure that all given quantities are converted to their standard SI units (meters, Coulombs, Farads, etc.) before substituting them into the formulas. This prevents common errors in calculation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves a parallel plate capacitor filled with a dielectric material. We need to find the dielectric constant (k) using the given physical quantities. The capacitance of a capacitor can be determined from its charge and voltage, and also from its physical dimensions and the dielectric material.

Step 2: Key Formula or Approach:
The capacitance \(C\) of a capacitor is given by the ratio of charge \(Q\) to the potential difference \(V\):
\[ C = \frac{Q}{V} \] For a parallel plate capacitor with a dielectric material, the capacitance is also given by:
\[ C = \frac{k \varepsilon_0 A}{d} \] where \(k\) is the dielectric constant, \(\varepsilon_0\) is the permittivity of free space (\(8.854 \times 10^{-12} \, \text{F/m}\)), \(A\) is the plate area, and \(d\) is the separation between the plates.
By equating these two expressions for \(C\), we can solve for \(k\).

Step 3: Detailed Explanation:
Given 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, \(V = 60 \, \text{V}\).
Calculation:
First, calculate the capacitance from the charge and voltage:
\[ C = \frac{Q}{V} = \frac{0.06 \times 10^{-6} \, \text{C}}{60 \, \text{V}} = 1 \times 10^{-9} \, \text{F} \] Next, rearrange the formula for the parallel plate capacitor to solve for the dielectric constant \(k\):
\[ k = \frac{C d}{\varepsilon_0 A} \] Now, substitute the known values into this equation:
\[ k = \frac{(1 \times 10^{-9} \, \text{F}) \times (2.0 \times 10^{-3} \, \text{m})}{(8.854 \times 10^{-12} \, \text{F/m}) \times (200 \times 10^{-4} \, \text{m}^2)} \] \[ k = \frac{2 \times 10^{-12}}{8.854 \times 10^{-12} \times 2 \times 10^{-2}} \] \[ k = \frac{2 \times 10^{-12}}{17.708 \times 10^{-14}} = \frac{2}{0.17708} \approx 11.294 \]

Step 4: Final Answer:
The calculated value of the dielectric constant is approximately 11.3.