Problem 39 - University Physics with Modern Physics

Question

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of \(1.60 \times\) \(10^{7} \mathrm{V} / \mathrm{m}\) . The capacitor is to have a capacitance of \(1.25 \times 10^{-9} \mathrm{F}\) and must be able to withstand a maximum potential difference of 5500 \(\mathrm{V} .\) What is the minimum area the plates of the capacitor may have?

Step-by-Step Solution

Answer

The minimum plate area is 135 cm².

Step 1: Understand the Problem

We need to find the minimum area of the plates for the given parallel-plate capacitor using its dielectric constant, dielectric strength, capacitance, and maximum potential difference.

Step 2: Use Capacitance Formula

The capacitance of a parallel-plate capacitor is given by \( C = \frac{\varepsilon_r \varepsilon_0 A}{d} \), where \( C \) is the capacitance, \( \varepsilon_r \) is the relative permittivity (dielectric constant), \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \, \mathrm{F/m}) \), \( A \) is the plate area, and \( d \) is the separation between the plates.

Step 3: Determine the Plate Separation

The dielectric strength \( (1.60 \times 10^{7} \, \mathrm{V/m}) \) gives us the maximum electric field \( (E) \) the capacitor can withstand. The maximum potential difference \( V \) is 5500 V, so the separation \( d \) can be found using \( d = \frac{V}{E} = \frac{5500}{1.60 \times 10^{7}} \).

Step 4: Calculate Plate Separation

Calculate \( d \) with the formula: \( d = \frac{5500}{1.60 \times 10^{7}} = 3.4375 \times 10^{-4} \, \mathrm{m} \).

Step 5: Solve for the Plate Area

We rearrange the capacitance formula to solve for the area \( A \): \( A = \frac{C \cdot d}{\varepsilon_r \varepsilon_0} \). Using this with \( C = 1.25 \times 10^{-9} \, \mathrm{F} \), \( \varepsilon_r = 3.60 \), and the previously calculated \( d \): \( A = \frac{1.25 \times 10^{-9} \times 3.4375 \times 10^{-4}}{3.60 \times 8.85 \times 10^{-12}} \).

Step 6: Calculate the Minimum Area

Calculate \( A \): \( A = \frac{1.25 \times 10^{-9} \times 3.4375 \times 10^{-4}}{3.186 \times 10^{-11}} = 0.0135 \, \mathrm{m^2} \) or \( 135 \, \mathrm{cm^2} \).

Step 7: Verify the Solution

Check if the calculated area allows the capacitor to meet the criteria by ensuring \( C = 1.25 \times 10^{-9} \mathrm{F} \) and can withstand \( 5500 \, \mathrm{V} \) under the maximum electric field condition. The calculation confirms the area is correct.

Key Concepts

Dielectric Constant

When discussing a parallel-plate capacitor, one crucial factor is the dielectric constant. A dielectric material is an insulating substance between the plates of a capacitor that increases its ability to store electrical energy. The dielectric constant, denoted as \( \varepsilon_r \), signifies how much the material can reduce the electric field compared to a vacuum, essentially representing its insulating efficiency.
This constant is dimensionless and always greater than 1, as it compares the permittivity of the dielectric material to that of a vacuum. For example, our exercise indicates a dielectric constant of 3.60, meaning the material is much more effective than a vacuum at storing energy.
Understanding the dielectric constant is crucial because it affects not only the capacitance but also the stability of the capacitor under different voltage applications.

Capacitance Formula

The capacitance formula for a parallel-plate capacitor provides a relationship between the physical and electrical properties of the capacitor. Capacitance \( C \) is defined by the equation:

• \( C = \frac{\varepsilon_r \varepsilon_0 A}{d} \)

where:

• \( \varepsilon_0 \) is the permittivity of free space \((8.85 \times 10^{-12} \, \text{F/m})\)
• \( A \) is the area of one of the capacitor's plates
• \( d \) is the separation between the plates

This formula is fundamental in assessing how changes in relative permittivity, plate area, or spacing influence the capacitor's ability to store charge.
It offers a direct approach to calculate the necessary dimensions of a capacitor based on the desired capacitance and the choice of dielectric used. In our scenario, we used it to determine the minimum plate area required to meet the specified capacitance of \(1.25 \times 10^{-9} \, \text{F}\).

Dielectric Strength

Dielectric strength is a critical property of the dielectric material used in capacitors. It represents the maximum electric field that a dielectric material can withstand without breaking down. Measured in volts per meter (V/m), it indicates the insulating power of the dielectric.
For instance, our problem states a dielectric strength of \(1.60 \times 10^{7} \, \text{V/m}\), meaning that beyond this limit, the material could fail, leading to a short circuit.
Determining the dielectric strength is important in ensuring the capacitor can handle the specified voltage. It provides engineers with a threshold to design safe and effective capacitors. In this exercise, we used dielectric strength to calculate the necessary separation between the plates to prevent breakdown at 5500 V.

Relative Permittivity

Relative permittivity, also known simply as permittivity or dielectric constant, is a key factor in determining a capacitor's efficiency. It quantifies how much electric displacement a dielectric material can generate when exposed to an electric field, compared to a vacuum.
This property directly influences the capacitance of a capacitor, as evident from the capacitance formula. Higher relative permittivity means a dielectric can allow more electric field lines to pass through, increasing storage capacity.
In calculating capacitance, relative permittivity is crucial because it accounts for the material's ability to affect the field between a capacitor's plates. Our exercise highlights this by using a \(\varepsilon_r \) of 3.60.

• This enhances the effectiveness of the capacitor compared to one with air as the dielectric.

Thus, selecting materials with the appropriate relative permittivity ensures that capacitors will perform effectively in their intended applications.