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An air capacitor is made from two flat parallel plates 1.50 mm apart. The magnitude of charge on each plate is 0.0180 \(\mu\)C when the potential difference is 200 V. (a) What is the capacitance? (b) What is the area of each plate? (c) What maximum voltage can be applied without dielectric breakdown? (Dielectric breakdown for air occurs at an electric-field strength of 3.0 \(\times\) 10\(^6\) V/m.) (d) When the charge is 0.0180 \(\mu\)C, what total energy is stored?

Short Answer

Expert verified
(a) 9.0 × 10^-11 F, (b) 0.0153 m², (c) 4500 V, (d) 1.8 × 10^-6 J

Step by step solution

01

Calculate the Capacitance

The capacitance of a parallel plate capacitor can be found using the formula \( C = \frac{Q}{V} \), where \( Q \) is the charge, and \( V \) is the potential difference. Given \( Q = 0.0180 \mu C = 0.0180 \times 10^{-6} \) C and \( V = 200 \) V, the capacitance \( C \) is calculated as:\[ C = \frac{0.0180 \times 10^{-6} \text{ C}}{200 \text{ V}} = 9.0 \times 10^{-11} \text{ F} \]
02

Calculate the Plate Area

The capacitance for a parallel plate capacitor is also given by the formula \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \text{ F/m}) \), \( A \) is the area, and \( d \) is the separation between plates. Rearrange the formula to find \( A \):\[ A = \frac{Cd}{\varepsilon_0} \]Substituting \( C = 9.0 \times 10^{-11} \text{ F} \), \( d = 1.50 \times 10^{-3} \text{ m} \), and \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m} \), we get:\[ A = \frac{9.0 \times 10^{-11} \times 1.50 \times 10^{-3}}{8.85 \times 10^{-12}} \approx 0.0153 \text{ m}^2 \]
03

Calculate Maximum Voltage for Dielectric Breakdown

The maximum voltage \( V_{\text{max}} \) that can be applied is calculated using the formula for electric field \( E = \frac{V}{d} \). Given the breakdown field strength \( E = 3.0 \times 10^6 \) V/m and \( d = 1.50 \times 10^{-3} \text{ m} \), solve for \( V_{\text{max}} \):\[ V_{\text{max}} = E \times d = 3.0 \times 10^6 \times 1.50 \times 10^{-3} = 4500 \text{ V} \]
04

Calculate the Total Energy Stored

The energy \( U \) stored in a capacitor is given by \( U = \frac{1}{2} C V^2 \). Using the previously calculated \( C = 9.0 \times 10^{-11} \text{ F} \) and \( V = 200 \text{ V} \), find \( U \):\[ U = \frac{1}{2} \times 9.0 \times 10^{-11} \times (200)^2 = 1.8 \times 10^{-6} \text{ J} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parallel Plate Capacitor
A parallel plate capacitor comprises two conductive plates separated by a certain distance. In this scenario, the plates are 1.50 mm apart. The capacitor stores electrical energy by maintaining opposite charges on the plates.

This setup enables the calculation of capacitance using the formula:
  • Capacitance, \( C \), is given by \( C = \frac{Q}{V} \) where \( Q \) is the charge on the plates and \( V \) is the voltage across them.
Capacitance is a measure of a capacitor's ability to store charge per unit voltage. For example, with a charge of 0.0180 \(\mu\)C and a voltage of 200 V, the capacitance of our parallel plate capacitor is 9.0 x 10\(^-{11}\ \text{F}\).

Additionally, capacitance can be calculated using the formula \( C = \frac{\varepsilon_0 A}{d} \), where:
  • \( \varepsilon_0 \) is the permittivity of free space, \( 8.85 \times 10^{-12} \text{ F/m} \)
  • \( A \) is the plate area
  • \( d \) is the distance separating the plates
This alternative method is particularly useful when you know the physical dimensions and need to connect them to how much charge can be stored.
Dielectric Breakdown
Dielectric breakdown is a critical phenomenon where the insulating material between capacitor plates fails to insulate, resulting in a sudden release of stored energy. This generally happens when a particular electric field strength is exceeded.

For air, the breakdown occurs at an electric field strength of 3.0 \( \times 10^6 \) V/m. Beyond this point, the air ceases to function as an effective insulator.

To ensure safe usage of a capacitor, it's crucial to determine the maximum voltage, \( V_{\text{max}} \), the capacitor can tolerate without breaking down. For the given example, it is calculated as:
  • \( V_{\text{max}} = E \times d \)
Inserting the air breakdown threshold \( E = 3.0 \times 10^6 \) V/m and the separation \( d = 1.50 \times 10^{-3} \) m yields a maximum voltage of 4500 V.
Electric Field Strength
Electric field strength is a measure of force per unit charge in the field around charged particles.It is denoted by \( E \) and is usually expressed in volts per meter (V/m).

The electric field between the plates of a parallel plate capacitor can be calculated using:
  • \( E = \frac{V}{d} \)
Where \( V \) is the potential difference across the plates and \( d \) is the distance between them.

In our example, maintaining an understanding of electric field strength helps avoid exceeding conditions for dielectric breakdown, as exceeding 3.0 \( \times 10^6 \) V/m will result in an electrical failure.

Electric field intensity is essential in designing capacitors ensuring they store energy efficiently without risking breakdown.
Energy Stored in a Capacitor
The energy stored in a capacitor is crucial for various applications where retraction of stored energy is needed. The energy is stored in the electric field between the plates, and the amount depends on both the capacitance and the square of the voltage.

The formula for energy stored, \( U \), in a capacitor is:
  • \( U = \frac{1}{2} C V^2 \)
Where:
  • \( C \) is the capacitance
  • \( V \) is the voltage across the capacitor
For the air capacitor with a charge of 0.0180 \mu C and a potential difference of 200 V, the energy stored turns out to be \( 1.8 \times 10^{-6} \) J.

This value portrays the potential of the capacitor as an energy reservoir - vital for applications like flash photography and power circuitry, where energy needs to be stored temporarily and then released rapidly.

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Most popular questions from this chapter

The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of 9.82 cm\(^2\). Each plate carries a charge of magnitude 4.35 \(\times\) 10\(^{-8}\) C. The plates are in vacuum. What is (a) the capacitance; (b) the potential difference between the plates; (c) the magnitude of the electric field between the plates?

A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF. (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?

A 12.5-\(\mu\)F capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease?

A cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to 125 V, \(\textbf{what is the charge per unit length}\) \(\lambda\) on the capacitor?

A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 mm. (a) If the sheets of paper measure 22 \(\times\)28 cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of 12.0 mm, instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Teflon than of posterboard? Explain.

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