Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \( V = 6.00 \, V \):\[ U = \frac{1}{2} (1.56 \times 10^{-12})(6.00^2) \approx 2.81 \times 10^{-11} \, J \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 20.0-\(\mu\)F capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to those of an uncharged 10.0-\(\mu\)F capacitor. Compute (a) the original charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 \(\times\) 10\(^7\) V>m. The capacitor is to have a capacitance of 1.25 \(\times\) 10\(^{-9}\) F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

Polystyrene has dielectric constant 2.6 and dielectric strength 2.0 \(\times\) 10\(^7\) V/m. A piece of polystyrene is used as a dielectric in a parallel-plate capacitor, filling the volume between the plates. (a) When the electric field between the plates is 80% of the dielectric strength, what is the energy density of the stored energy? (b) When the capacitor is connected to a battery with voltage 500.0 V, the electric field between the plates is 80% of the dielectric strength. What is the area of each plate if the capacitor stores 0.200 mJ of energy under these conditions?

You have two identical capacitors and an external potential source. (a) Compare the total energy stored in the capacitors when they are connected to the applied potential in series and in parallel. (b) Compare the maximum amount of charge stored in each case. (c) Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?

A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to potentials of up to 1.00 \(\times\) 10\(^2\) V. The electric field between the plates is to be no greater than 1.00 \(\times\) 10\(^4\) N/C. As a budding electrical engineer for Live-Wire Electronics, your tasks are to (a) design the capacitor by finding what its physical dimensions and separation must be; (b) find the maximum charge these plates can hold.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free