Problem 2 - University Physics with Modern Physics

Question

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

(a) \(3.29 \times 10^{-12} \, \mathrm{F}\); (b) \(13218.8 \, \mathrm{V}\); (c) \(4.03 \times 10^{6} \, \mathrm{V/m}\).

Step 1: Convert units if necessary

First, convert the given dimensions into SI units. The distance between the plates is given as 3.28 mm, which we convert to meters: \(d = 3.28 \times 10^{-3} \mathrm{m}\). The area is given as 12.2 cm², which we convert to square meters: \(A = 12.2 \times 10^{-4} \mathrm{m}^2\).

Step 2: Find the capacitance (a)

The capacitance of a parallel-plate capacitor is given by the formula \(C = \frac{\varepsilon_0 A}{d}\), where \(\varepsilon_0\) is the vacuum permittivity, approximately \(8.85 \times 10^{-12} \, \mathrm{F/m}\). Substituting the values, we have: \[C = \frac{8.85 \times 10^{-12} \, \mathrm{F/m} \times 12.2 \times 10^{-4} \, \mathrm{m}^2}{3.28 \times 10^{-3} \, \mathrm{m}} \approx 3.29 \times 10^{-12} \, \mathrm{F}.\]

Step 3: Calculate the potential difference (b)

The potential difference between the plates is calculated using the charge and capacitance: \(V = \frac{Q}{C}\). Using \(Q = 4.35 \times 10^{-8} \, \mathrm{C}\) and the capacitance found in Step 2:\[V = \frac{4.35 \times 10^{-8} \, \mathrm{C}}{3.29 \times 10^{-12} \, \mathrm{F}} \approx 13218.8 \, \mathrm{V}.\]

Step 4: Compute the magnitude of the electric field (c)

The electric field \(E\) between the plates is related to the potential difference and the distance by \(E = \frac{V}{d}\). From Step 3, \(V \approx 13218.8 \, \mathrm{V}\) and \(d = 3.28 \times 10^{-3} \, \mathrm{m}\):\[E = \frac{13218.8 \, \mathrm{V}}{3.28 \times 10^{-3} \, \mathrm{m}} \approx 4.03 \times 10^{6} \, \mathrm{V/m}.\]

Key Concepts

Capacitance Calculation

Calculating the capacitance of a parallel-plate capacitor is straightforward if you understand the core formula. The capacitance, denoted by \( C \), tells you how much charge the capacitor can store per unit voltage. For a parallel-plate capacitor, this is calculated using:
\[ C = \frac{\varepsilon_0 A}{d} \]
Here's what each symbol means:

• \( \varepsilon_0 \) - Vacuum permittivity, a constant approximately equal to \( 8.85 \times 10^{-12} \, \text{F/m} \), representing how much electric field the vacuum can "permit".


• \( A \) - Area of one of the plates, which needs to be in square meters \( \text{m}^2 \).


• \( d \) - Distance between the two plates, in meters \( \text{m} \).

To calculate the capacitance:

• Convert all provided units to \( \text{meters} \) and \( \text{square meters} \).


• Plug these values into the formula.


• The result is the capacitance \( C \), measured in Farads \( \text{F} \).

By applying the given values, we find the capacitance to be approximately \( 3.29 \times 10^{-12} \, \text{F} \), indicating how much charge the capacitor can hold per volt.

Potential Difference

In the context of a parallel-plate capacitor, the potential difference, or voltage \( V \), is the difference in electric potential between the two plates. This can be calculated using:
\[ V = \frac{Q}{C} \]
Where:

• \( Q \) is the charge on one plate, measured in Coulombs \( \text{C} \).


• \( C \) is the capacitance calculated earlier.

For our capacitor:

• The charge \( Q \) is given as \( 4.35 \times 10^{-8} \, \text{C} \).


• The capacitance \( C \) is approximately \( 3.29 \times 10^{-12} \, \text{F} \).

By dividing \( Q \) by \( C \), we find the potential difference to be about \( 13218.8 \, \text{V} \). This indicates the energy per unit charge exerted by the electric field across the capacitor.

Electric Field

The electric field \( E \) between the plates of a parallel-plate capacitor is an important concept, representing the force on a charge per unit charge in that field. It can be determined via the equation:
\[ E = \frac{V}{d} \]
In this formula:

• \( V \) is the potential difference across the plates, measured in Volts \( \text{V} \).


• \( d \) is the distance between the plates, in meters \( \text{m} \).

Taking our earlier results:

• \( V \approx 13218.8 \, \text{V} \).


• \( d = 3.28 \times 10^{-3} \, \text{m} \).

Substituting these values, the electric field \( E \) comes out to approximately \( 4.03 \times 10^{6} \, \text{V/m} \). This measure tells us about the intensity of the field, showing how strongly a charge will be acted upon across the space between the plates.