Question

A parallel plate capacitor has a capacitance of \(C=10 \mathrm{pF}\). The plates have area \(0.025 \mathrm{~cm}^{2}\). A dielectric layer of thickness \(d_{\text {thick }}=0.01 \mathrm{~mm}\) separates the plates. For the dielectric layer, calculate the permittivity \(\epsilon,\) the relative permittivity \(\epsilon_{r},\) and the electric susceptibility \(\chi_{e}\)

Short answer

\( \epsilon = 4 \times 10^{-11} \text{ F/m}, \epsilon_{r} \approx 4.52, \chi_{e} = 3.52 \)

Step-by-step solution

Identify the formula for capacitance with a dielectric

The formula for the capacitance of a parallel plate capacitor with a dielectric is given by \[ C = \epsilon \frac{A}{d} \]where:- \( C \) is the capacitance (10 pF),- \( A \) is the area of the plates (0.025 cm² = 2.5 \times 10^{-6} \text{ m}^2),- \( d \) is the separation between the plates (0.01 mm = 1 \times 10^{-5} \text{ m}),- \( \epsilon \) is the permittivity of the dielectric.

Solve for permittivity \(\epsilon\)

Rearrange the formula to solve for \( \epsilon \):\[ \epsilon = \frac{C \cdot d}{A} \]Substitute the given values:\[ \epsilon = \frac{10 \times 10^{-12} \text{ F} \cdot 1 \times 10^{-5} \text{ m}}{2.5 \times 10^{-6} \text{ m}^2} = 4 \times 10^{-11} \text{ F/m} \]

Calculate the relative permittivity \(\epsilon_{r}\)

The relative permittivity \( \epsilon_{r} \) is given by:\[ \epsilon_{r} = \frac{\epsilon}{\epsilon_0} \]where \( \epsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \) is the permittivity of free space.\[ \epsilon_{r} = \frac{4 \times 10^{-11} \text{ F/m}}{8.85 \times 10^{-12} \text{ F/m}} \approx 4.52 \]

Calculate the electric susceptibility \(\chi_{e}\)

The electric susceptibility \( \chi_{e} \) is related to the relative permittivity as:\[ \chi_{e} = \epsilon_{r} - 1 \]From step 3:\[ \chi_{e} = 4.52 - 1 = 3.52 \]

Key concepts

Dielectric

A dielectric is a non-conducting material that is placed between the plates of a capacitor to increase its capacitance. Unlike conductive materials that allow electric charges to flow freely, dielectrics are insulators that hold their electric charges in place. When a dielectric is placed in an electric field, it becomes polarized. This means that the dielectric particles align themselves according to the field, with positive and negative charges leaning in opposite directions.

The introduction of a dielectric increases the capacitance, which allows the capacitor to store more electric charge at the same voltage level. This enhancement happens because the electric field inside the dielectric is weaker than in a vacuum due to its polarization. Consequently, the net field between the plates reduces, which decreases the voltage for the same charge on the plates. Therefore, more charge can be stored for a given voltage.

Permittivity

Permittivity, often denoted by the symbol \( \epsilon \), is a measure that describes how much electric field is reduced in a medium compared to a vacuum. It represents the ability of a dielectric material to permit electric field lines to pass through it. Permittivity is given in the units of Farads per meter (F/m).

There are two types of permittivities:

• **Absolute Permittivity (\( \epsilon \))**: This refers to the actual permittivity of the material in question. It directly determines how much the material can reduce the electric field between the capacitor plates. In our problem, we calculated \( \epsilon = 4 \times 10^{-11} \text{ F/m} \).
• **Relative Permittivity (\( \epsilon_r \))**: This is the ratio of the permittivity of a material to the permittivity of free space (\( \epsilon_0 \)), which is \( 8.85 \times 10^{-12} \text{ F/m} \). It shows how much more effective the dielectric is compared to a vacuum. For instance, \( \epsilon_r \approx 4.52 \) in our exercise.

Relative permittivity provides valuable insights into how good a material is at storing electrical energy within an electric field. A greater \( \epsilon_r \) implies a stronger dielectric.

Electric Susceptibility

Electric susceptibility, symbolized as \( \chi_e \), is another important property of dielectrics. It measures how easily a material can be polarized by an external electric field. Practically, it shows how much the dielectric can "align" its internal charges in response to an electric field, thereby enhancing the material's electric field reducing capability.

Electric susceptibility is related to the relative permittivity of a material through the formula: \( \chi_e = \epsilon_r - 1 \). This relation helps in understanding the extent to which a material can become polarized compared to free space. In our problem, \( \chi_e \approx 3.52 \), indicating that the material significantly affects the electric field inside it.

By knowing both \( \epsilon_r \) and \( \chi_e \), we can determine the strength and effectiveness of dielectrics. High values of electric susceptibility indicate that a material is effective in reducing the electric field, thereby enhancing the capacitance of capacitors in practical applications.