Ch 24: Gauss' Law

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 24: Gauss' Law

Problem 21

Chapter 24, Problem 21

What is the net electric flux through the cylinder of FIGURE EX24.21?

Verified step by step guidance

Identify the relevant law: The net electric flux through a closed surface is given by Gauss's Law, which states:

Φ = \frac{q (enclosed)}{ε}

, where

q

is the total charge enclosed by the surface and

ε

is the permittivity of free space.

Determine the charge enclosed by the cylinder: Examine the figure (FIGURE EX24.21) to identify the charges inside the cylindrical surface. Sum up all the charges enclosed by the cylinder to find

q

Substitute the value of the enclosed charge into Gauss's Law: Use the formula

Φ = \frac{q (enclosed)}{ε}

to calculate the net electric flux. Ensure that the charge is in Coulombs and the permittivity of free space is

8.85 \times 10^{-12} F \cdot m^{-1}

Account for symmetry: If the cylinder is symmetric and the electric field is uniform, the flux through the curved surface may cancel out, leaving only contributions from the ends of the cylinder. Verify this based on the figure.

Conclude the net flux: After substituting the values and considering symmetry, the net electric flux through the cylinder is determined by the total enclosed charge divided by the permittivity of free space. Ensure all units are consistent throughout the calculation.

Verified video answer for a similar problem:

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

Video duration:

Was this helpful?

Key Concepts

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

Gauss's Law

Gauss's Law states that the net electric flux through a closed surface is proportional to the charge enclosed within that surface. Mathematically, it is 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 principle is fundamental for calculating electric fields in symmetrical charge distributions.

Electric Flux

Electric flux is a measure of the electric field passing through a given area. It is defined as the product of the electric field (E) and the area (A) through which it passes, adjusted for the angle (θ) between the field lines and the normal to the surface: Φ_E = E · A · cos(θ). Understanding electric flux is crucial for applying Gauss's Law effectively.

Symmetry in Charge Distributions

Symmetry in charge distributions simplifies the calculation of electric fields and flux. For example, in cylindrical or spherical geometries, the electric field can be assumed uniform over certain surfaces, allowing for straightforward application of Gauss's Law. Recognizing the symmetry helps in determining the enclosed charge and thus the net electric flux through the surface.

Recommended video:

Guided course

12:39

Solving Symmetric Launch Problems

Related Practice

Textbook Question

Find the electric fluxes Φ_A to Φ_E through surfaces A to E in FIGURE P24.29.

views

Textbook Question

A 2.0 cm × 3.0 cm rectangle lies in the $x z$ -plane with unit vector $\hat{n}$ pointing in the +y-direction. What is the electric flux through the rectangle if the electric field is $\vec{E} = (4000 \hat{i} - 2000 \hat{k})$ N/C?

views

Textbook Question

A 12 cm × 12 cm rectangle lies in the first quadrant of the xy-plane with one corner at the origin. Unit vector $\hat{n}$ points in the +𝒵-direction. What is the electric flux through the rectangle if the electric field is $E = (2000 m^{- 1}) x \hat{k}$ N/C? Hint: Divide the rectangle into narrow strips of width.

views

Textbook Question

FIGURE EX24.18 shows three charges. Draw these charges on your paper four times. Then draw two-dimensional cross sections of three-dimensional closed surfaces through which the electric flux is (a) $2 q divided by epsilon sub 0$ , (b) $q divided by epsilon sub 0$ , (c) 0, and (d) $5 q divided by epsilon sub 0$ .

Textbook Question

A thin, horizontal, 10-cm-diameter copper plate is charged to 3.5 nC. If the charge is uniformly distributed on the surface, what are the strength and direction of the electric field 0.1 mm above the center of the top surface of the plate?

Textbook Question

A spark occurs at the tip of a metal needle if the electric field strength exceeds $3.0 \times 10^{6}$ N/C, the field strength at which air breaks down. What is the minimum surface charge density for producing a spark?

views