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Chapter 17: Problem 100

A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\). The parallel plates are separated by \(0.40 \mathrm{mm}\) of air. (a) What is the capacitance for this capacitor? (b) What is the area of a single plate? (c) At what voltage will the air between the plates become ionized? Assume a dielectric strength of \(3.0 \mathrm{kV} / \mathrm{mm}\) for air.

Short Answer

Expert verified
Answer: The capacitance of the capacitor is \(8.33 \times 10^{-11} \text{F}\), the area of a single plate is \(3.76 \times 10^{-3} \text{m\)^2\(}\), and the voltage at which the air between the plates becomes ionized is \(1.2 \text{kV}\).

Step by step solution

01

Find the capacitance

Using the formula for the capacitance of a parallel plate capacitor: \(C = \frac{Q}{V}\) Where \(C\) is the capacitance, \(Q\) is the charge, and \(V\) is the potential difference. We have \(Q = 0.020 \mu \mathrm{C}\) and \(V = 240 \mathrm{V}\). Plug these values in the formula and solve for \(C\): \(C = \frac{0.020 \times 10^{-6} \text{C}}{240 \text{V}}\)
02

Calculate the capacitance

Calculate the capacitance: \(C = \frac{0.020 \times 10^{-6} \text{C}}{240 \text{V}} = 8.33 \times 10^{-11} \text{F}\) The capacitance of this capacitor is \(8.33 \times 10^{-11} \text{F}\).
03

Find the area of a single plate

To find the area of a single plate, we use the formula for the capacitance of a parallel plate capacitor with a dielectric constant: \(C = \frac{\epsilon_0 A}{d}\) Where \(C\) is the capacitance, \(\epsilon_0\) is the vacuum permittivity \(8.85 \times 10^{-12} \text{F/m}\), \(A\) is the area of a single plate, and \(d\) is the separation distance between the plates. We have \(C = 8.33 \times 10^{-11} \text{F}\) and \(d = 0.40 \mathrm{mm}\). Plug these values into the formula and solve for \(A\): \(A = \frac{C \times d}{\epsilon_0} = \frac{8.33 \times 10^{-11} \text{F} \times 0.40 \times 10^{-3} \text{m}}{8.85 \times 10^{-12} \text{F/m}}\)
04

Calculate the area of a single plate

Calculate the area of a single plate: \(A = \frac{8.33 \times 10^{-11} \text{F} \times 0.40 \times 10^{-3} \text{m}}{8.85 \times 10^{-12} \text{F/m}} = 3.76 \times 10^{-3} \text{m\)^2\(}\) The area of a single plate is \(3.76 \times 10^{-3} \text{m\)^2\(}\).
05

Find the voltage at which the air becomes ionized

Given the dielectric strength of air is \(3.0 \text{kV/mm}\), we can find the voltage at which the air becomes ionized by multiplying the dielectric strength by the separation distance: \(V_{ionization} = \text{Dielectric strength} \times \text{Separation distance}\) We have the dielectric strength as \(3.0 \text{kV/mm}\) and the separation distance as \(0.40 \mathrm{mm}\). Multiply them to get: \(V_{ionization} = 3.0 \text{kV/mm} \times 0.40 \mathrm{mm}\)
06

Calculate the ionization voltage

Calculate the ionization voltage: \(V_{ionization} = 3.0 \text{kV/mm} \times 0.40 \mathrm{mm} = 1.2 \text{kV}\) The voltage at which the air between the plates becomes ionized is \(1.2 \text{kV}\). To summarize the results: a) The capacitance of the capacitor is \(8.33 \times 10^{-11} \text{F}\). b) The area of a single plate is \(3.76 \times 10^{-3} \text{m\)^2\(}\). c) The voltage at which the air between the plates becomes ionized is \(1.2 \text{kV}\).

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Most popular questions from this chapter

A parallel plate capacitor is connected to a battery. The space between the plates is filled with air. The electric field strength between the plates is \(20.0 \mathrm{V} / \mathrm{m} .\) Then, with the battery still connected, a slab of dielectric \((\kappa=4.0)\) is inserted between the plates. The thickness of the dielectric is half the distance between the plates. Find the electric field inside the dielectric.
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