Problem 1 - University Physics with Modern Physics

Question

The plates of a parallel-plate capacitor are 2.50 \(\mathrm{mm}\) apart, and each carries a charge of magnitude 80.0 \(\mathrm{nC}\) . The plates are in vacuum. The electric field between the plates has a magnitude of \(4.00 \times 10^{6} \mathrm{V} / \mathrm{m}\) . (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the capacitance?

Step-by-Step Solution

Answer

(a) 10000 V; (b) 1.77 x 10^-2 m^2; (c) 8.00 x 10^-12 F.

Step 1: Calculate the Potential Difference

The potential difference (abla V) between the plates can be calculated using the formula for the electric field (E): abla V = E \cdot d, where \( d \) is the distance between the plates. Given \( E = 4.00 \times 10^{6} \, \mathrm{V/m} \) and \( d = 2.50 \times 10^{-3} \, \mathrm{m} \), we can substitute these values to find the potential difference: \( abla V = 4.00 \times 10^{6} \, \mathrm{V/m} \times 2.50 \times 10^{-3} \, \mathrm{m} = 10000 \, \mathrm{V} \).

Step 2: Calculate the Capacitance

The capacitance (C) of a parallel-plate capacitor is given by the formula C = \( \frac{Q}{abla V} \), where Q is the charge and \( abla V \) is the potential difference. We already calculated \( abla V = 10000 \, \mathrm{V} \) and \( Q = 80.0 \times 10^{-9} \, \mathrm{C} \). Substituting these values into the formula gives \( C = \frac{80.0 \times 10^{-9} \, \mathrm{C}}{10000 \, \mathrm{V}} = 8.00 \times 10^{-12} \, \mathrm{F} \).

Step 3: Calculate the Area of Each Plate

For a parallel-plate capacitor, the electric field is also given by E = \( \frac{\sigma}{\varepsilon_0} \), where \( \sigma \) is the surface charge density and \( \varepsilon_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} \, \mathrm{F/m} \)). \( \sigma \) is expressed as \( \frac{Q}{A} \), where \( A \) is the area of the plate. We have E and need A: \( A = \frac{Q \varepsilon_0}{E} \). Substituting \( Q = 80.0 \times 10^{-9} \), \( \varepsilon_0 = 8.85 \times 10^{-12} \), and \( E = 4.00 \times 10^6 \), we find: \( A = \frac{80.0 \times 10^{-9} \times 8.85 \times 10^{-12}}{4.00 \times 10^6} = 1.77 \times 10^{-2} \mathrm{m^2} \).

Key Concepts

Capacitance

Capacitance is a fundamental property of a capacitor, which measures its ability to store an electric charge. When considering capacitors, it tells us how much charge, denoted by \( Q \), can be stored per unit of potential difference, \( \Delta V \), across its plates. The formula for capacitance is given by:

• \( C = \frac{Q}{\Delta V} \)

It is measured in Farads (F), where one Farad is equal to one Coulomb per Volt.
In our case, a parallel-plate capacitor exhibits capacitance based on the charge it holds and the potential difference between its plates. With a charge of \( 80.0 \, \text{nC} \) and a measured potential difference of \( 10000 \, \text{V} \), we can calculate the capacitance to be \( 8.00 \times 10^{-12} \, \text{F} \), or 8 picoFarads.
Capacitance is a critical concept in designing circuits as it affects how a circuit responds to voltage changes, storing energy until it's needed.

Electric Field

The electric field is a region around charged particles or objects within which electric forces are exerted on other charges. In simpler terms, it's the force per unit charge experienced by a small positive test charge placed in the field. The electric field strength, \( E \), is expressed in volts per meter (V/m). For parallel-plate capacitors, the field is uniform between the plates and is determined by:

• \( E = \frac{\Delta V}{d} \)

where \( \Delta V \) is the potential difference and \( d \) is the distance between the plates.
In the given problem, the electric field between the plates is \( 4.00 \times 10^6 \, \text{V/m} \), demonstrating a strong field even with just a small separation of \( 2.50 \, \text{mm} \). Thus, when capacitors are used in a vacuum, this uniform field can achieve a higher force over a small area.

Potential Difference

Potential difference, also known as voltage, is the work done to move a charge from one point to another in an electric field. It's a crucial element in electric circuits because it's responsible for driving charges through the circuit, allowing electrical devices to operate.
It is calculated as:

• \( \Delta V = E \cdot d \)

For the parallel-plate capacitor example, with an electric field \( E = 4.00 \times 10^6 \, \text{V/m} \) and plate separation \( d = 2.50 \times 10^{-3} \, \text{m} \), the potential difference is found to be \( 10000 \, \text{V} \) (or 10 kV).
Understanding the potential difference between two points helps us in analyzing energy conversion and transfer in capacitive systems.

Parallel-Plate Capacitor

The parallel-plate capacitor is one of the simplest and most commonly used forms of capacitors. It comprises two conductive plates separated by a small distance, often filled with a dielectric material, although in a vacuum for our problem.
The primary formula for the capacitance of a parallel-plate capacitor is given by:

• \( C = \varepsilon_0 \frac{A}{d} \)

where \( \varepsilon_0 \) is the vacuum permittivity (\( 8.85 \times 10^{-12} \, \text{F/m} \)), \( A \) is the plate area, and \( d \) is the separation between the plates.
In solving this problem, we determine the plate area to be \( 1.77 \times 10^{-2} \, \text{m}^2 \). This design appeals for storing energy efficiently in electric fields generated between the plates. Parallel-plate capacitors are integral in devices for smoothing electrical signals and energy storage.