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Chapter 26: Problem 47

A parallel plate capacitor with \(C=0.050 \mu \mathrm{F}\) has a separation between its plates of \(d=50.0 \mu \mathrm{m} .\) The dielectric that fills the space between the plates has dielectric constant \(\kappa=2.5\) and resistivity \(\rho=4.0 \cdot 10^{12} \Omega \mathrm{m} .\) What is the time constant for this capacitor? (Hint: First calculate the area of the plates for the given \(C\) and \(\kappa\), and then determine the resistance of the dielectric between the plates.)

Short Answer

Expert verified
The time constant of the capacitor is approximately 8500 seconds.

Step by step solution

01

Calculate the area of the plates

To calculate the area of the plates, we should first rearrange the formula for capacitance: \(A = \frac{Cd}{\kappa \epsilon_0}\). We are given \(C = 0.050 \mu \mathrm{F}\), \(d = 50.0 \mu \mathrm{m}\), and \(\kappa = 2.5\). We also know that the vacuum permittivity constant \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}\). Plugging in these values, we get: $$A = \frac{(0.050 \times 10^{-6} \mathrm{F})(50.0 \times 10^{-6} \mathrm{m})}{2.5 (8.85 \times 10^{-12} \mathrm{F/m})} \approx 1.12 \times 10^{-3} \mathrm{m^2}$$
02

Calculate the resistance of the dielectric

Now, we need to calculate the resistance \(R\) of the dielectric using the formula \(R = \frac{\rho L}{A}\). We are given the resistivity \(\rho = 4.0 \times 10^{12} \Omega \mathrm{m}\) and the separation between the plates \(d = 50.0 \mu \mathrm{m} = L\). We have found the area \(A = 1.12 \times 10^{-3} \mathrm{m^2}\). Plugging in these values, we get: $$R = \frac{(4.0 \times 10^{12} \Omega \mathrm{m})(50.0 \times 10^{-6} \mathrm{m})}{1.12 \times 10^{-3} \mathrm{m^2}} \approx 1.7 \times 10^8 \Omega$$
03

Calculate the time constant

Finally, we can calculate the time constant \(\tau\) using the formula \(\tau = RC\). We have found \(R = 1.7 \times 10^8 \Omega\) and we were given \(C = 0.050 \mu \mathrm{F}\). Plugging in these values, we get: $$\tau = (1.7 \times 10^8 \Omega)(0.050 \times 10^{-6} \mathrm{F}) \approx 8500~\mathrm{s}$$ The time constant for this capacitor is approximately 8500 seconds.

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Most popular questions from this chapter

Many electronics devices can be dangerous even after they are shut off. Consider an RC circuit with a \(150 .-\mu \mathrm{F}\) capacitor and a \(1.00-\mathrm{M} \Omega\) resistor connected to a \(200 .-\mathrm{V}\) power source for a long time and then disconnected and shorted, as shown in the figure. How long will it be until the potential difference across the capacitor drops to below \(50.0 \mathrm{~V} ?\)

The dead battery of your car provides a potential difference of \(9.950 \mathrm{~V}\) and has an internal resistance of \(1.100 \Omega\). You charge it by connecting it with jumper cables to the live battery of another car. The live battery provides a potential difference of \(12.00 \mathrm{~V}\) and has an internal resistance of \(0.0100 \Omega\), and the starter resistance is \(0.0700 \Omega\). a) Draw the circuit diagram for the connected batteries. b) Determine the current in the live battery, in the dead battery, and in the starter immediately after you closed the circuit.

A battery, a resistor, and a capacitor are connected in series in an RC circuit. What happens to the current through a resistor after a long time? Explain using Kirchhoff's rules.

The ammeter your physics instructor uses for in-class demonstrations has internal resistance \(R_{\mathrm{i}}=75 \Omega\) and measures a maximum current of \(1.5 \mathrm{~mA}\). The same ammeter can be used to measure currents of much greater magnitudes by wiring a shunt resistor of relatively small resistance, \(R_{\text {shunt }}\), in parallel with the ammeter. (a) Sketch the circuit diagram, and explain why the shunt resistor connected in parallel with the ammeter allows it to measure larger currents. (b) Calculate the resistance the shunt resistor has to have to allow the ammeter to measure a maximum current of 15 A.

How long would it take, in multiples of the time constant, \(\tau,\) for the capacitor in an \(\mathrm{RC}\) circuit to be \(98 \%\) charged? a) \(9 \tau\) c) \(90 \tau\) e) \(0.98 \tau\) b) \(0.9 \tau\) d) \(4 \tau\)
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