Question

Two parallel plate capacitors, \(C_{1}\) and \(C_{2},\) are con nected in series with a \(60.0-\mathrm{V}\) battery and a \(300 .-\mathrm{k} \Omega\) resistor, as shown in the figure. Both capacitors have plates with an area of \(2.00 \mathrm{~cm}^{2}\) and a separation of \(0.100 \mathrm{~mm}\). Capacitor \(C_{1}\) has air between its plates, and capacitor \(C_{2}\) has the gap filled with a certain porcelain (dielec-

Short answer

Question: In a series circuit of a battery (DC voltage, V = 60 V), a resistor, \(C_1\) (a capacitor with air between its plates), \(C_2\) (a capacitor with a dielectric constant \(k = 6.0\)), determine (a) the magnitude of the capacitive reactance of the circuit, (b) the magnitude of the charge on each capacitor, and (c) the magnitude of the charge on the resistor. Given: Area of the capacitor plates, \(A = 2.00 \mathrm{~cm}^2\), and the separation between the plates, \(d = 0.100 \mathrm{~mm}\). Answer: (a) The magnitude of the capacitive reactance is \(\infty \mathrm{~\Omega}\), (b) the magnitude of the charge on each capacitor is \(87.78 \times 10^{-6} \mathrm{C}\), and (c) the magnitude of the charge on the resistor is \(0 \mathrm{C}\).

Step-by-step solution

Find the capacitance of \(C_2\)

To find the capacitance of \(C_2\), we will use the formula \(C = k \epsilon_0 \frac{A}{d}\). The area \(A = 2.00 \mathrm{~cm}^2 = 0.0002 \mathrm{~m}^2\) and the separation \(d = 0.100 \mathrm{~mm} = 10^{-4} \mathrm{~m}\). We have \(\epsilon_0 \approx 8.85 \times 10^{-12} \mathrm{F/m}\). Then, the capacitance of \(C_2\) is \(C_2 = 6.0 \times 8.85 \times 10^{-12} \times \frac{0.0002}{10^{-4}} \mathrm{F} = 10.62 \times 10^{-9} \mathrm{F}\).

Find the capacitance of \(C_1\)

Since \(C_1\) has air between its plates, we have \(k=1\). Therefore, the capacitance of \(C_1\) is \(C_1 = \epsilon_0 \frac{A}{d} = 8.85 \times 10^{-12} \times \frac{0.0002}{10^{-4}} \mathrm{F} = 1.77 \times 10^{-9} \mathrm{F}\).

Find the equivalent capacitance

Since the capacitors are connected in series, the equivalent capacitance \(C_{eq}\) can be found using the formula \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\). Plugging in the values, we get \(\frac{1}{C_{eq}} = \frac{1}{1.77 \times 10^{-9}} + \frac{1}{10.62 \times 10^{-9}}\). Solving for \(C_{eq}\), we get \(C_{eq} = 1.463 \times 10^{-9} \mathrm{F}\).

Calculate the capacitive reactance

To calculate the capacitive reactance, we will use the formula \(X_c = \frac{1}{ωC_{eq}}\), where \(ω\) is the angular frequency. Since the battery is DC, the capacitive reactance will be \(X_c = \infty \mathrm{~\Omega}\) (meaning no current will flow after the capacitors are fully charged).

Calculate the charge on each capacitor

The charge \(Q\) on a charged capacitor can be obtained using the formula \(Q = C_\text{eq} V\), where \(V=60.0 \mathrm{V}\) is the battery voltage. Solving for \(Q\), we get \(Q = 1.463 \times 10^{-9} \mathrm{F} \times 60.0 \mathrm{V} = 87.78 \times 10^{-6} \mathrm{C}\). Since the capacitors are connected in series, they store the same charge. Therefore, the charge on \(C_1\) and \(C_2\) is \(Q_{C_1} = Q_{C_2} = 87.78 \times 10^{-6} \mathrm{C}\).

Calculate the charge on the resistor

Since the capacitors are fully charged and no current is flowing through the circuit, the charge on the resistor will be \(Q_{R} = 0 \mathrm{C}\). In conclusion, the capacitive reactance of the circuit is \(\infty \mathrm{~\Omega}\), the charge on each capacitor is \(87.78 \times 10^{-6} \mathrm{C}\), and the charge on the resistor is \(0 \mathrm{C}\).

Key concepts

Parallel Plate Capacitors

Parallel plate capacitors are foundational components in electronics. They consist of two conductive plates situated parallel to one another, separated by a small distance. This structure allows them to store electric charge on their plates. The ability of a parallel plate capacitor to store charge is defined by its capacitance, denoted as \( C \).
Capacitance depends on several factors:

• Area (\( A \)) of the plates: Larger plates can hold more charge, increasing capacitance.
• Distance (\( d \)) between the plates: A smaller distance increases the electric field's strength and thus the capacitance.
• Dielectric material between the plates: This material can affect capacitance, as seen with the dielectric constant discussed next.

For those dealing with capacitors in practical use, understanding these factors is crucial. They allow one to manage how much charge a capacitor can store, making them indispensable in various applications such as energy storage and signal processing.

Dielectric Constant

Dielectric constant, represented as \( k \), plays a critical role in capacitors' functionality. It measures a material's ability to reduce the electric field inside the capacitor, thereby allowing it to store more charge.
When a dielectric material, such as porcelain, is inserted between capacitor plates:

• The capacitance increases, as is evident in our case with \( C_2 \).
• Capacitance formula becomes \( C = k \epsilon_0 \frac{A}{d} \), illustrating the multiplier effect of \( k \) on the capacitor's ability to store charge.
• The dielectric constant of air, often used as a reference, is 1.0.

By implementing materials with higher dielectric constants, engineers can enhance the performance of capacitors significantly, selected based on the application needs, for example, in tuning circuits and memory storage. Understanding this property is vital for designing optimal capacitive systems.

Series Circuit

Capacitors can be connected in a series configuration, which significantly influences the overall circuitry. In a series circuit, components are linked end-to-end, forming a single path for current flow.
Important characteristics of series circuits:

• In a series configuration, the total or equivalent capacitance \( C_{eq} \) is less than any individual capacitance and is calculated as \( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots \).
• This has a particular effect on voltage distribution—each capacitor will have a different voltage drop depending on its individual capacitance.
• However, they share the same charge when fully charged, simplifying calculations for distributed charge, as shown with \( C_1 \) and \( C_2 \) in the provided solution.

Understanding the behavior in series is essential for designing circuits where voltage division and charge distribution need careful management, such as in voltage multipliers.

Charge on Capacitors

The charge on capacitors in any circuit is a fundamental property, directly illustrating their function—the ability to store and release electric energy.
Some key points about the charge on capacitors are:

• Charge \( Q \) on a capacitor is determined by the equation \( Q = CV \), where \( C \) is capacitance, and \( V \) is the voltage applied.
• In series, as discussed earlier, the same charge is stored across all capacitors—an essential aspect for circuits designed for equal charge distribution.
• Understanding how much charge a capacitor holds is crucial for predicting how it will interact within a larger circuit, affecting functionality and efficiency.

Charge management is essential for applications ranging from simple electronic gadgets to complex power systems, aligning the capacitor's role as an energy store and release component effectively for optimal performance.