Table of contents

0. Math Review31m
- Math Review
  31m
1. Intro to Physics Units1h 29m
2. 1D Motion / Kinematics3h 56m
3. Vectors2h 43m
4. 2D Kinematics1h 42m
5. Projectile Motion3h 6m
6. Intro to Forces (Dynamics)3h 22m
7. Friction, Inclines, Systems2h 44m
8. Centripetal Forces & Gravitation7h 26m
9. Work & Energy1h 59m
10. Conservation of Energy2h 54m
11. Momentum & Impulse3h 40m
12. Rotational Kinematics2h 59m
13. Rotational Inertia & Energy7h 4m
14. Torque & Rotational Dynamics2h 5m
15. Rotational Equilibrium3h 39m
16. Angular Momentum3h 6m
17. Periodic Motion2h 9m
18. Waves & Sound3h 40m
19. Fluid Mechanics4h 27m
20. Heat and Temperature3h 7m
21. Kinetic Theory of Ideal Gases1h 50m
22. The First Law of Thermodynamics1h 26m
23. The Second Law of Thermodynamics3h 11m
24. Electric Force & Field; Gauss' Law3h 42m
25. Electric Potential1h 51m
26. Capacitors & Dielectrics2h 2m
27. Resistors & DC Circuits3h 8m
28. Magnetic Fields and Forces2h 23m
29. Sources of Magnetic Field2h 30m
30. Induction and Inductance3h 38m
31. Alternating Current2h 37m
32. Electromagnetic Waves2h 14m
33. Geometric Optics2h 57m
34. Wave Optics1h 15m
35. Special Relativity2h 10m

24. Electric Force & Field; Gauss' Law

Electric Field

Physics

24. Electric Force & Field; Gauss' Law

Electric Field

Previous problemNext problem

2:19 minutes

Problem 23

Textbook Question

A hollow copper sphere has inner radius 1.0 cm and outer radius 2.5 cm. A 5.0 A current flows radially outward from the inner surface to the outer surface. What is the electric field strength at r=2.0 cm?

Verified step by step guidance

Step 1: Understand the problem. The electric field strength at a distance r from the center of the hollow sphere can be calculated using the relationship between current density, charge flow, and electric field. The current flows radially outward, and the geometry of the sphere suggests that the electric field depends on the radial distance r.

Step 2: Calculate the current density J. The current density is defined as the current per unit area. For a spherical shell, the area at a radial distance r is given by the surface area of a sphere: $ A = 4 \pi r^2 $. Use $ J = \frac{I}{A} $, where $ I $ is the current and $ A $ is the area.

Step 3: Relate the current density to the electric field. In a conductor, the current density $ J $ is related to the electric field $ E $ by the equation $ J = \sigma E $, where $ \sigma $ is the conductivity of the material. Rearrange this equation to solve for $ E $: $ E = \frac{J}{\sigma} $.

Step 4: Determine the conductivity $ \sigma $ of copper. Look up the conductivity of copper (a known material property) or use the value provided in the problem if specified. Substitute $ \sigma $ and $ J $ into the equation for $ E $.

Step 5: Evaluate $ E $ at $ r = 2.0 $ cm. Substitute $ r = 2.0 $ cm into the expressions for $ A $ and $ J $, and then calculate $ E $ using the formula derived in the previous steps. Ensure all units are consistent (e.g., convert cm to meters).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field

The electric field is a vector field that represents the force exerted by an electric charge on other charges in its vicinity. It is defined as the force per unit charge and is measured in volts per meter (V/m). In this scenario, the electric field strength at a specific radius within the hollow sphere is crucial for understanding how the current affects the surrounding space.

Ohm's Law

Ohm's Law states that the current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. This relationship is expressed as V = IR. In the context of the hollow sphere, understanding how current relates to electric field and resistance is essential for calculating the electric field strength.

Gauss's Law

Gauss's Law relates the electric flux passing through a closed surface to the charge enclosed by that surface. It is mathematically expressed as Φ_E = Q_enc/ε_0, where Φ_E is the electric flux, Q_enc is the enclosed charge, and ε_0 is the permittivity of free space. This law is particularly useful in this problem to determine the electric field at a point within the hollow sphere by considering the symmetry and the distribution of the current.

Watch next

Master Intro to Electric Fields with a bite sized video explanation from Patrick

Start learning

Related Videos

Related Practice

Textbook Question

The electric field at a point in space is $E = (400 \hat{i} + 100 \hat{j})$ N/C. What is the electric force on a proton at this point? Give your answer in component form.

1003

views

Textbook Question

The nuclei of large atoms, such as uranium, with $92$ protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately $7.4\times10^{-15}$ m. What magnitude of electric field does it produce at the distance of the electrons, which is about $1.0 \times 10^{- 10}$ m?

862

views

Textbook Question

A Van de Graaff generator is a device for generating a large electric potential by building up charge on a hollow metal sphere. A typical classroom-demonstration model has a diameter of 30 cm. What is the electric field strength just outside the surface of the sphere when it is charged to 500,000 V?

689

views

Textbook Question

What are the strength and direction of the electric field 1.0 mm from a proton?

709

views

Textbook Question

Two point charges, Q₁ = - 32 μC and Q₂ = +45μC, are separated by a distance of 12 cm. The electric field at the point P (see Fig. 21–61) is zero. How far from Q₁ is P?

971

views

Textbook Question

You are given two unknown point charges, Q₁ and Q₂. At a point on the line joining them, one-third of the way from Q₁ to Q₂, the electric field is zero (Fig. 21–64). What is the ratio Q₁/Q₂?

597

views

Textbook Question

(II) A positive charge q is placed at the center of a circular ring of radius R. The ring carries a uniformly distributed negative charge of total magnitude -Q. (a) If the charge q is displaced from the center a small distance x as shown in Fig. 21–71, show that it will undergo simple harmonic motion when released. (b) If its mass is m, what is its period?

664

views

Textbook Question

A point charge of mass 0.145 kg, and net charge +3.40 μC, hangs at rest at the end of an insulating cord above a large sheet of charge. The horizontal sheet of fixed uniform charge creates a uniform vertical electric field in the vicinity of the point charge. The tension in the cord is measured to be 5.18 N. (a) Calculate the magnitude and direction of the electric field due to the sheet of charge (Fig. 21–80). (b) What is the surface charge density σ(C/m²) on the sheet?

344

views

Show more practices

Download the Mobile app

Privacy

Do not sell my personal information

Accessibility

Patent notice

Sitemap