Problem 37 - College Physics

Question

\(\bullet\) The plates of a parallel-plate capacitor are 3.28 \(\mathrm{mm}\) apart, and each has an area of 12.2 \(\mathrm{cm}^{2} .\) Each plate carries a charge of magnitude \(4.35 \times 10^{-8} \mathrm{C}\) . The plates are in vacuum. (a) What is the capacitance? (b) What is the potential difference between the plates? (c) What is the magnitude of the electric field between the plates?

Step-by-Step Solution

Answer

Capacitance: 3.29 pF; Potential difference: 13,224 V; Electric field: 4.03x10^6 V/m.

Step 1: Convert Units

First, let's convert the units given in the problem to standard SI units. - Convert the plate separation from millimeters to meters: \[3.28 \text{ mm} = 3.28 \times 10^{-3} \text{ m}\] - Convert the area from square centimeters to square meters: \[12.2 \text{ cm}^2 = 12.2 \times 10^{-4} \text{ m}^2\]

Step 2: Calculate the Capacitance

The formula for the capacitance of a parallel-plate capacitor in a vacuum is:\[C = \varepsilon_0 \frac{A}{d}\]where - \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}\) (permittivity of free space),- \(A = 12.2 \times 10^{-4} \text{ m}^2\) (area of the plates),- \(d = 3.28 \times 10^{-3} \text{ m}\) (distance between the plates).Substitute the given values:\[C = 8.85 \times 10^{-12} \text{ F/m} \times \frac{12.2 \times 10^{-4} \text{ m}^2}{3.28 \times 10^{-3} \text{ m}}\]\[C \approx 3.29 \times 10^{-12} \text{ F}\]Thus, the capacitance is approximately 3.29 pF.

Step 3: Find the Potential Difference

The potential difference \(V\) between the plates of a capacitor is related to the charge \(Q\) and the capacitance \(C\) by the formula:\[V = \frac{Q}{C}\]where - \(Q = 4.35 \times 10^{-8} \text{ C}\), - \(C = 3.29 \times 10^{-12} \text{ F}\).Substitute the values:\[V = \frac{4.35 \times 10^{-8} \text{ C}}{3.29 \times 10^{-12} \text{ F}}\]\[V \approx 13223.71 \text{ V}\]Thus, the potential difference is approximately 13,224 V.

Step 4: Calculate the Electric Field

The magnitude of the electric field \(E\) between the plates is given by:\[E = \frac{V}{d}\]where - \(V \approx 13223.71 \text{ V}\) - \(d = 3.28 \times 10^{-3} \text{ m}\).Substitute the values:\[E = \frac{13223.71 \text{ V}}{3.28 \times 10^{-3} \text{ m}}\]\[E \approx 4.03 \times 10^6 \text{ V/m}\]Thus, the magnitude of the electric field between the plates is approximately \(4.03 \times 10^6 \text{ V/m}\).

Key Concepts

Parallel-Plate Capacitor

A parallel-plate capacitor is a simple electrical component composed of two flat, conductive plates facing each other. These plates are separated by a small distance, usually filled with a dielectric material like a vacuum, air, or other insulating substances. In our specific problem, the plates are 3.28 mm apart and placed in a vacuum.
The primary role of a parallel-plate capacitor is to store electrical energy in an electric field that is established between the plates when they hold opposite but equal charges. Due to this simple setup, these capacitors are widely used for theoretical studies to understand basic electrical properties and are essential in various practical applications like tuning circuits and energy harvesting devices.
The effectiveness of a parallel-plate capacitor is determined by specific factors, such as the area of the plates and the distance between them. Interestingly, increasing the plate area or decreasing the distance of separation between the plates increases the capacitor's ability to store charge. This is quantified by a property known as capacitance, which we'll calculate in later sections.

Electric Field

The electric field is a vector field around charged particles, and it is especially relevant between the plates of a parallel-plate capacitor. It's created by the charge present on these plates and points from the positive to the negative plate.
In our scenario, each plate has specific charge amounts, and this, combined with the separation distance, allows calculation of the electric field. The strength of this field is consistent between two parallel plates and can be calculated using the formula:

• \(E = \frac{V}{d}\)

where \( V \) is the potential difference and \( d \) is the distance between the plates.
Understanding the electric field is vital because it represents the force that would act on a positive test charge placed between the plates. It is measured in volts per meter (V/m), and its magnitude in our case is approximately \( 4.03 \times 10^6 \) V/m. This number indicates a strong field that simplifies the movement of charges between the plates, contributing to the efficiency of the capacitor.

Potential Difference

The potential difference, also known as voltage, is crucial in understanding the operation of a capacitor. It defines how much work is required to move a charge between two points—in this case, between the plates of the capacitor.
For a parallel-plate capacitor, the potential difference is linked with the charge it stores and its capacitance. We can calculate this value using the formula:

• \(V = \frac{Q}{C}\)

where \(Q\) is the charge stored (4.35 x 10^{-8} C in this scenario) and \(C\) is the capacitance, which we calculated earlier as approximately 3.29 pF.
After performing the calculation with these values, the potential difference is approximately 13,224 volts. This tells us that a high amount of energy is required to separate the charges onto the two plates when a capacitor maintains such a high voltage, making it a powerful component for storing energy.

Capacitance Calculation

Capacitance is a measure of a capacitor's ability to store charge per unit voltage across its plates. It depends on the physical characteristics of the capacitor, such as the surface area of the plates and the separation distance between them.
For a parallel-plate capacitor, the capacitance \( C \) is given by the formula:

• \(C = \varepsilon_0 \frac{A}{d}\)

where \(\varepsilon_0\) is the permittivity of free space, \(A\) is the area of the plates (converted to square meters in our calculation), and \(d\) is the distance between the plates.
Substituting the known values, we find that the capacitance is approximately 3.29 picofarads (pF). This calculated value indicates how much electric charge the capacitor can hold for each volt of charge applied to it, offering insights into its storage potential once the capacitor is part of an electrical circuit.