Problem 72 - College Physics

Question

The paper dielectric in a paper-and-foil capacitor is 0.0800 mm thick. Its dielectric constant is \(2.50,\) and its dielectric strength is 50.0 \(\mathrm{MV} / \mathrm{m}\) . Assume that the geometry is that of a parallel- plate capacitor, with the metal foil serving as the plates. (a) What area of each plate is required for a 0.200\(\mu F\) capacitor? (b) If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the capacitor?

Step-by-Step Solution

Answer

(a) Plate area: \(7.25 \times 10^{-2} \text{ m}^2\). (b) Max potential difference: 2000 V.

Step 1: Identify Given Values

We are given the thickness of the dielectric as \( d = 0.0800 \text{ mm} = 0.0800 \times 10^{-3} \text{ m} \). The dielectric constant \( \kappa = 2.50 \), dielectric strength \( E_{\text{max}} = 50.0 \text{ MV/m} \), and the desired capacitance \( C = 0.200 \mu F = 0.200 \times 10^{-6} \text{ F} \).

Step 2: Use Capacitance Formula

The formula for capacitance with a dielectric is \( C = \frac{\kappa \varepsilon_0 A}{d} \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the thickness of the dielectric.

Step 3: Rearrange for Plate Area

Rearrange the formula to solve for the plate area \( A = \frac{C \cdot d}{\kappa \varepsilon_0} \).

Step 4: Calculate Plate Area

Substitute the known values into the equation: \( A = \frac{0.200 \times 10^{-6} \text{ F} \cdot 0.0800 \times 10^{-3} \text{ m}}{2.50 \cdot 8.85 \times 10^{-12} \text{ F/m}} \approx 7.25 \times 10^{-2} \text{ m}^2 \).

Step 5: Determine Maximum Electric Field

The maximum electric field under half the dielectric strength condition is \( E = \frac{E_{\text{max}}}{2} = \frac{50.0 \text{ MV/m}}{2} = 25.0 \text{ MV/m} \).

Step 6: Calculate Maximum Voltage

The maximum potential difference \( V \) is given by \( V = E \cdot d \). Substitute in the values: \( V = 25.0 \times 10^{6} \text{ V/m} \cdot 0.0800 \times 10^{-3} \text{ m} = 2000 \text{ V} \).

Key Concepts

Dielectric Constant

The dielectric constant, often symbolized as \( \kappa \), is a measure of a material's ability to increase the capacitance of a capacitor compared to a vacuum. It essentially quantifies how effectively a material can store electrical energy. In simpler terms, think of it as the material's ability to "boost" the effect of the capacitor beyond what would be possible without it.
Key points about the dielectric constant include:

• A higher dielectric constant means the material can store more electrical charge.
• Materials with higher dielectric constants are often used to create capacitors with better performance.
• In our specific problem, the dielectric constant value given is \( 2.50 \), indicating that the paper in the capacitor makes it 2.5 times more effective than if it were just a vacuum.

Understanding the concept of dielectric constant is crucial for calculating how much charge a capacitor can store with a given dielectric material.

Dielectric Strength

Dielectric strength refers to the maximum electric field that a material can withstand without breaking down or losing its insulating properties. It's essentially a measure of the electrical strength of a material as an insulator.
Some essential aspects to remember about dielectric strength include:

• It is measured in volts per meter (V/m).
• The higher the dielectric strength, the better the material is at resisting electrical breakdown.
• For our paper dielectric, the dielectric strength is given as \( 50.0 \text{ MV/m} \). This means it can resist an electric field up to this strength before failing.
• To avoid breakdown, it is wise to operate below the maximum dielectric strength, as shown in the problem where the maximum field allowed is half the dielectric strength, \( 25.0 \text{ MV/m} \).

Dielectric strength is crucial for determining the safety and stability of capacitors under high voltages.

Parallel-Plate Capacitor

A parallel-plate capacitor is a simple type used frequently in electronics. It consists of two conductive plates separated by an insulating material, known as the dielectric.
Characteristics of parallel-plate capacitors include:

• They have straightforward geometry, making them easy to analyze mathematically.
• The capacitance depends on the surface area of the plates, the distance between the plates, and the dielectric material used.
• In our problem, the geometry is assumed to be that of a parallel-plate capacitor, where the metal foils act as plates sandwiching the paper dielectric.

This simple yet effective design is why parallel-plate capacitors are commonly used for teaching and in practical applications.

Capacitance Formula

The formula for the capacitance \( C \) of a capacitor, particularly with a dielectric, is essential for calculating the amount of charge it can store. The formula is:\[ C = \frac{\kappa \varepsilon_0 A}{d} \]where:

• \( \kappa \) is the dielectric constant, representing the material's ability to store charge.
• \( \varepsilon_0 \) is the permittivity of free space, a constant value of \( 8.85 \times 10^{-12} \) F/m.
• \( A \) is the area of one of the plates.
• \( d \) is the separation between the plates, or the thickness of the dielectric.

In the given problem, this formula helps determine the area of the plates needed to achieve the desired capacitance of \( 0.200 \mu F \). Understanding how to manipulate this formula is vital for designing capacitors that fit specific electrical requirements. By rearranging it, you can find any of the components, given the rest.