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

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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

01

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 \(C_2 = 6.0 \times 8.85 \times 10^{-12} \times \frac{0.0002}{10^{-4}} \mathrm{F} = 10.62 \times 10^{-9} \mathrm{F}\).
02

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 \(C_1 = \epsilon_0 \frac{A}{d} = 8.85 \times 10^{-12} \times \frac{0.0002}{10^{-4}} \mathrm{F} = 1.77 \times 10^{-9} \mathrm{F}\).
03

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 \(C_{eq} = 1.463 \times 10^{-9} \mathrm{F}\).
04

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).
05

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 \(Q_{C_1} = Q_{C_2} = 87.78 \times 10^{-6} \mathrm{C}\).
06

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 \(0 \mathrm{C}\).

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