Problem 39 - College Physics

Question

A parallel-plate air capacitor has a capacitance of 500.0 \(\mathrm{pF}\) and a charge of magnitude 0.200\(\mu \mathrm{C}\) on each plate. The plates are 0.600 \(\mathrm{mm}\) apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electric-field magnitude between the plates? (d) What is the surface charge density on each plate?

Step-by-Step Solution

Answer

(a) 400 V, (b) 0.0339 m², (c) 666,667 V/m, (d) 5.90 µC/m².

Step 1: Find the Potential Difference

The potential difference between the plates of a capacitor can be found using the formula \( V = \frac{Q}{C} \), where \( Q \) is the charge and \( C \) is the capacitance. Here, \( Q = 0.200 \times 10^{-6} \) C, and \( C = 500.0 \times 10^{-12} \) F. Thus, \( V = \frac{0.200 \times 10^{-6}}{500.0 \times 10^{-12}} = 400 \text{ volts} \).

Step 2: Determine the Area of Each Plate

The area of each plate can be determined using the formula for capacitance in a parallel-plate capacitor: \( C = \frac{\varepsilon_0 A}{d} \), where \( A \) is the area, \( \varepsilon_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} \text{ F/m} \)), and \( d \) is the separation distance. Rearranging gives \( A = \frac{C \cdot d}{\varepsilon_0} \). Plugging in, \( A = \frac{500.0 \times 10^{-12} \times 0.600 \times 10^{-3}}{8.85 \times 10^{-12}} \approx 0.0339 \text{ m}^2 \).

Step 3: Calculate the Electric-field Magnitude

The electric-field magnitude \( E \) between the plates of a capacitor is given by \( E = \frac{V}{d} \). From Step 1, \( V = 400 \text{ volts} \) and \( d = 0.600 \times 10^{-3} \text{ m} \). So, \( E = \frac{400}{0.600 \times 10^{-3}} = 666,667 \text{ V/m} \).

Step 4: Compute the Surface Charge Density

The surface charge density \( \sigma \) is given by \( \sigma = \frac{Q}{A} \). Using \( Q = 0.200 \times 10^{-6} \text{ C} \) and the area found in Step 2, \( A = 0.0339 \text{ m}^2 \), \( \sigma = \frac{0.200 \times 10^{-6}}{0.0339} \approx 5.90 \times 10^{-6} \text{ C/m}^2 \).

Key Concepts

Parallel-Plate Capacitor

A parallel-plate capacitor is a device that stores electrical energy. It consists of two conductive plates separated by a small distance, with an insulating material, such as air, in between. The key feature of a parallel-plate capacitor is its simple design, which makes it easy to analyze and understand.
The capacitance of a parallel-plate capacitor is determined by its physical characteristics and is defined by the formula:

• \(C = \frac{\varepsilon_0 A}{d}\)
  • \(C\) = Capacitance (in Farads)
  • \(\varepsilon_0\) = Permittivity of free space \(\approx 8.85 \times 10^{-12} \text{ F/m}\)
  • \(A\) = Area of one plate (in square meters)
  • \(d\) = Distance between the plates (in meters)

By connecting these components in a circuit, the capacitor can be charged, storing energy in the electric field between the plates. This simple structure makes the parallel-plate capacitor a foundational component in many electronic devices.

Electric Field

The electric field in a parallel-plate capacitor is a critical component that determines how the device operates. It is a vector field between the plates that exerts force on electric charges, and can be calculated using the formula:

• \(E = \frac{V}{d}\)
• \(E\) = Electric field (in volts per meter)
• \(V\) = Potential difference between plates (in volts)
• \(d\) = Separation distance between the plates (in meters)

The electric field is uniform and points from the positive to the negative plate, helping to hold the energy within the capacitor. This uniformity is critical for calculating other properties of the capacitor. Knowing the electric field allows us to further explore how the capacitor will interact in any given circuit, predicting the behavior of other components affected by its presence.

Surface Charge Density

In the context of a parallel-plate capacitor, surface charge density is important because it tells you how much charge is concentrated on a unit area of the capacitor plates. You can find the surface charge density with the formula:

• \(\sigma = \frac{Q}{A}\)
• \(\sigma\) = Surface charge density (in coulombs per square meter)
• \(Q\) = Total charge on one plate (in coulombs)
• \(A\) = Area of the plate (in square meters)

Surface charge density gives insight into how well a capacitor might perform. Higher density means more charge is stored for a given area, allowing the capacitor to store more energy overall. This is vital for applications requiring high energy or where space is limited, as it directly impacts efficiency.

Capacitance

Capacitance is the property of the parallel-plate capacitor that defines its ability to store an electrical charge. It's a measure of how much charge a capacitor can hold at a given potential difference across its plates, and it's expressed in farads (F).

• For a parallel-plate capacitor:
  • \(C = \frac{Q}{V}\)
  • \(Q\) = Charge on the plates (in coulombs)
  • \(V\) = Voltage between the plates (in volts)
• The formula \(C = \frac{\varepsilon_0 A}{d}\) also applies, highlighting the physical factors affecting capacitance.

A higher capacitance means a capacitor can store more charge at a lower voltage, which is beneficial for various electronic applications. Understanding capacitance is crucial for designing circuits that require energy storage, filtering, or timing elements, as it describes how capacitors will interact with the rest of the circuit.