Understanding the Parallel Plate Capacitor Problem

This question asks us to find the dielectric constant ($k$) of the material placed between the plates of a parallel plate capacitor. We are given the plate area ($A$), the separation between the plates ($d$), the charge ($Q$) stored on the capacitor, and the potential difference ($V$) across the plates.

Given Values and Unit Conversions

First, let's list the given values and convert them into SI units, which are necessary for our calculations:

• Plate Area, $A = 200 \text{ cm}^2$. To convert to square meters ($m^2$): $A = 200 \times (10^{-2} \text{ m})^2 = 200 \times 10^{-4} \text{ m}^2 = 2.0 \times 10^{-2} \text{ m}^2$.
• Separation between plates, $d = 2.0 \text{ mm}$. To convert to meters ($m$): $d = 2.0 \times 10^{-3} \text{ m}$.
• Charge on the capacitor, $Q = 0.06 \mu\text{C}$. To convert to Coulombs ($C$): $Q = 0.06 \times 10^{-6} \text{ C}$.
• Potential difference, $V = 60 \text{ V}$. This is already in SI units.
• Permittivity of free space, $\epsilon_0 \approx 8.854 \times 10^{-12} \text{ F/m}$ (Farads per meter). This is a physical constant.

Calculating Capacitance

The relationship between charge ($Q$), capacitance ($C$), and potential difference ($V$) for any capacitor is given by the formula:

We can rearrange this formula to find the capacitance ($C$) using the given charge and voltage:

Substituting the given values:

So, the capacitance of the parallel plate capacitor is $1 \times 10^{-9}$ Farads.

Determining the Dielectric Constant

The capacitance of a parallel plate capacitor filled with a dielectric material is given by the formula:

Where:

• $k$ is the dielectric constant of the material.
• $\epsilon_0$ is the permittivity of free space.
• $A$ is the area of one plate.
• $d$ is the separation between the plates.

We need to find the dielectric constant ($k$). We can rearrange the formula to solve for $k$:

Step-by-Step Calculation of Dielectric Constant

Now, we substitute the calculated capacitance ($C$) and the given/converted values of $d$, $\epsilon_0$, and $A$ into the formula for $k$:

$k = \frac{(1 \times 10^{-9} \text{ F}) \times (2.0 \times 10^{-3} \text{ m})}{(8.854 \times 10^{-12} \text{ F/m}) \times (2.0 \times 10^{-2} \text{ m}^2)}$

Let's calculate the numerator and the denominator separately:

• Numerator: $(1 \times 10^{-9}) \times (2.0 \times 10^{-3}) = 2.0 \times 10^{-12} \text{ F} \cdot \text{m}$
• Denominator: $(8.854 \times 10^{-12}) \times (2.0 \times 10^{-2}) = 17.708 \times 10^{-14} \text{ F} \cdot \text{m}$

Now, divide the numerator by the denominator:

$k = \frac{2.0 \times 10^{-12}}{17.708 \times 10^{-14}} = \frac{2.0}{17.708} \times \frac{10^{-12}}{10^{-14}}$

$k = \frac{2.0}{17.708} \times 10^{-12 - (-14)} = \frac{2.0}{17.708} \times 10^{2}$

$k \approx 0.11294 \times 100 \approx 11.294$

Final Result

The calculated value for the dielectric constant ($k$) is approximately 11.3. This value corresponds to the material filled between the capacitor plates.