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A 0.350-m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 mm and an outer hollow conducting tube with an inner radius of 2.00 mm. The two conductors are separated by air and charged to a potential difference of 6.00 V. Calculate (a) the charge per length for the capacitor; (b) the total charge on the capacitor; (c) the capacitance; (d) the energy stored in the capacitor when fully charged.

Short Answer

Expert verified
(a) Charge per length \( \lambda \approx 2.68 \times 10^{-11} \, C/m \). (b) Total charge \( Q \approx 9.38 \times 10^{-12} \, C \). (c) Capacitance \( C \approx 1.56 \times 10^{-12} \, F \). (d) Energy stored \( U \approx 2.81 \times 10^{-11} \, J \).

Step by step solution

01

Identify the formula for voltage difference (V) in a cylindrical capacitor

In a cylindrical capacitor, the electric potential difference \( V \) between the core and outer tube is given by the formula \[ V = \frac{\lambda}{2\pi\varepsilon_0} \ln\left(\frac{r_2}{r_1}\right) \] where \( \lambda \) is the charge per unit length, \( \varepsilon_0 \) is the permittivity of free space \((\approx 8.85 \times 10^{-12} \, C^2/N\cdot m^2)\), \( r_1 \) is the radius of the inner core, and \( r_2 \) is the radius of the outer hollow conductor.
02

Calculate charge per length (\( \lambda \))

Rearrange the formula from Step 1 to solve for charge per unit length \( \lambda \): \[ \lambda = \frac{2\pi\varepsilon_0 V}{\ln(r_2/r_1)} \] Substitute \( V = 6.00 \, V \), \( r_1 = 1.20 \times 10^{-3} \, m \), \( r_2 = 2.00 \times 10^{-3} \, m \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \, C^2/N\cdot m^2 \):\[ \lambda = \frac{2\pi (8.85 \times 10^{-12})(6.00)}{\ln(2.00/1.20)} \approx 2.68 \times 10^{-11} \, C/m \].
03

Calculate the total charge (Q) on the capacitor

The total charge \( Q \) is the charge per unit length \( \lambda \) times the length of the capacitor \( L \):\[ Q = \lambda \times L \] Given \( L = 0.350 \, m \), and using the \( \lambda \) calculated:\[ Q = (2.68 \times 10^{-11} \, C/m)(0.350 \, m) \approx 9.38 \times 10^{-12} \, C \].
04

Determine the capacitance (C)

The capacitance of a cylindrical capacitor is given by:\[ C = \frac{2\pi\varepsilon_0 L}{\ln(r_2/r_1)} \]Substitute the given values \( L = 0.350 \, m \), \( r_1 = 1.20 \times 10^{-3} \, m \), \( r_2 = 2.00 \times 10^{-3} \, m \), and \( \varepsilon_0 = 8.85 \times 10^{-12} \, C^2/N\cdot m^2 \):\[ C = \frac{2\pi (8.85 \times 10^{-12})(0.350)}{\ln(2.00/1.20)} \approx 1.56 \times 10^{-12} \, F \].
05

Calculate the energy stored (U)

The energy stored in the capacitor when fully charged can be calculated using:\[ U = \frac{1}{2} C V^2 \]Substitute \( C = 1.56 \times 10^{-12} \, F \) and \( V = 6.00 \, V \):\[ U = \frac{1}{2} (1.56 \times 10^{-12})(6.00^2) \approx 2.81 \times 10^{-11} \, J \].

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Key Concepts

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

Charge per Length in a Cylindrical Capacitor
In the context of capacitors, especially cylindrical ones, it is important to understand the term "charge per length." This refers to how much electric charge is distributed along a given length of the cylindrical surface. The symbol \( \lambda \) is commonly used to represent this quantity.

The cylindrical capacitor features a solid core and an outer hollow tube. To find the charge per length \( \lambda \), we use the formula:
  • \[ \lambda = \frac{2\pi\varepsilon_0 V}{\ln(r_2/r_1)} \]
In this formula:
  • \( \varepsilon_0 \) signifies the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, C^2/N\cdot m^2 \).
  • \( V \) is the potential difference between the core and tube.
  • \( r_1 \) and \( r_2 \) are the radii of the inner core and outer tube, respectively.
To solve an example from a given problem, you might substitute specific values such as \( V = 6.00 \, V \), \( r_1 = 1.20 \times 10^{-3} \, m \), and \( r_2 = 2.00 \times 10^{-3} \, m \). This gives a charge per length \[ \lambda \approx 2.68 \times 10^{-11} \, C/m \].

Understanding charge per length lays a foundation for calculating total charge and other properties of capacitors.
Capacitance of a Cylindrical Capacitor
Capacitance is a measure of the ability of a capacitor to store an electrical charge. In cylindrical capacitors, it depends on the geometry and material properties. For a cylindrical capacitor, the formula for capacitance \( C \) is:
  • \[ C = \frac{2\pi\varepsilon_0 L}{\ln(r_2/r_1)} \]
Here:
  • \( L \) is the length of the cylindrical capacitor.
  • Again, \( r_1 \) and \( r_2 \) are the radii of the inner and outer parts, respectively.
  • \( \varepsilon_0 \) is the permittivity of free space, as before.
Using our example values, such as \( L = 0.350 \, m \), you calculate the capacitance to be approximately \[ C \approx 1.56 \times 10^{-12} \, F \].

The capacitance indicates how much charge the capacitor can hold when a particular voltage is applied. This makes capacitance a central quantity in the analysis of electrical circuits.
Energy Stored in a Capacitor
The energy stored in a capacitor is crucial for understanding how capacitors function in circuits. This stored energy can be calculated using the formula:
  • \[ U = \frac{1}{2} C V^2 \]
Where:
  • \( C \) is the capacitance found earlier, \( 1.56 \times 10^{-12} \, F \) in our drumroll example.
  • \( V \) is the potential difference, given as \( 6.00 \, V \).
Substituting these values into the formula, the energy stored \( U \) is \[ 2.81 \times 10^{-11} \, J \].

This energy is essentially the potential energy that can be released into a circuit to do work, such as lighting a bulb or powering a device. It highlights the utility of capacitors in storing and releasing energy quickly when needed.

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Most popular questions from this chapter

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00\(\times\) 10\(^6\) V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 10.0 pC. The inner cylinder has radius 0.50 mm, the outer one has radius 5.00 mm, and the length of each cylinder is 18.0 cm. (a) What is the capacitance? (b) What applied potential difference is necessary to produce these charges on the cylinders?

When a 360-nF air capacitor (11 nF = 10\(^{-9}\) F) is connected to a power supply, the energy stored in the capacitor is 1.85 \(\times\) 10\(^{-5}\) J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.32 \(\times\) 10\(^{-5}\) J. (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?

In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, airfilled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of each metal plate is 42.0 mm\(^2\), and the separation between the plates is 0.700 mm before the key is depressed. (a) Calculate the capacitance before the key is depressed. (b) If the circuitry can detect a change in capacitance of 0.250 pF, how far must the key be depressed before the circuitry detects its depression?

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?

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