Ch 23: The Electric Field

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo

Knight Calc 5th Edition

Ch 23: The Electric Field

Problema 41

Capitolo 23, Problema 41

FIGURE P23.41 is a cross section of two infinite lines of charge that extend out of the page. Both have linear charge density λ. Find an expression for the electric field strength E at height y above the midpoint between the lines.

Guida verificata passo dopo passo

Identify the setup: The problem involves two infinite lines of charge with linear charge density λ. The electric field due to an infinite line of charge is radially outward (or inward for negative charge) and decreases with distance. The goal is to find the net electric field at a point located at height y above the midpoint between the lines.

Write the expression for the electric field due to a single infinite line of charge: The magnitude of the electric field at a distance r from an infinite line of charge is given by the formula:

E = \frac{1}{2 π ε ₀} \frac{λ}{r}

, where λ is the linear charge density, r is the perpendicular distance from the line, and ε₀ is the permittivity of free space.

Determine the geometry: Let the two lines of charge be separated by a distance d. The point of interest is at height y above the midpoint. The distance from the point to each line of charge can be calculated using the Pythagorean theorem:

r = \sqrt{y^{2} + {\frac{d}{2}}^{2}}

Resolve the electric field components: The electric field due to each line of charge has both horizontal and vertical components. The horizontal components (x-direction) will cancel out because the lines are symmetrically placed, while the vertical components (y-direction) will add up. The vertical component of the electric field from one line is:

E_y = \frac{1}{2 π ε ₀} \frac{λ}{r} \frac{y}{r}

, where

\frac{y}{r}

accounts for the vertical component.

Combine the contributions: Since there are two lines of charge, the total electric field in the y-direction is twice the vertical component from one line:

E = 2 \frac{1}{2 π ε ₀} \frac{λ}{r} \frac{y}{r}

. Substitute

r = \sqrt{y^{2} + {\frac{d}{2}}^{2}}

to express the electric field in terms of y, d, and λ.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Electric Field Due to a Line of Charge

The electric field generated by an infinite line of charge is directed radially outward (or inward, depending on the charge sign) and decreases with distance from the line. The magnitude of the electric field E at a distance r from a line with linear charge density λ is given by E = (λ / (2πε₀r)), where ε₀ is the permittivity of free space. This concept is crucial for calculating the resultant electric field at a point due to multiple lines of charge.

Video consigliato:

Percorso guidato

08:20

Electric Field Lines

Superposition Principle

The superposition principle states that the total electric field at a point due to multiple charge distributions is the vector sum of the electric fields produced by each charge distribution individually. In this case, the electric fields from both lines of charge must be calculated separately and then combined to find the net electric field at the specified height y above the midpoint.

Video consigliato:

Percorso guidato

03:32

Superposition of Sinusoidal Wave Functions

Coordinate System and Geometry

Understanding the geometry of the problem is essential for correctly applying the principles of electric fields. In this scenario, the height y and the distance from each line of charge must be accurately represented in a coordinate system. The symmetry of the configuration allows for simplifications in calculations, particularly when determining the direction and magnitude of the electric fields from each line of charge.

Video consigliato:

Percorso guidato

05:17

Coordinates of Center of Mass of 4 objects

Pratica correlata

Domanda del libro di testo

Derive Equation 23.11 for the field Ē dipole in the plane that bisects an electric dipole.

195

views

Domanda del libro di testo

What are the strength and direction of the electric field at the position indicated by the dot in FIGURE P23.37? Give your answer (a) in component form and (b) as a magnitude and angle measured cw or ccw (specify which) from the positive x-axis.

210

views

Domanda del libro di testo

A −15 nC charge is at x=+2.0 cm on the x-axis. A second charge q is located somewhere on the x-axis to the left of the origin. The electric field at y=2.0 cm on the y-axis is $\vec{E} = 3.0 \times 10^{5} \hat{i}$ N/C . What are (a) the charge q in nC and (b) its distance from the origin?

2589

views

Domanda del libro di testo

FIGURE P23.44 shows a thin rod of length L with total charge Q. Evaluate E at r=3.0 cm if L=5.0 cm and Q=3.0 nC.

views

Domanda del libro di testo

A ring of radius R has total charge Q. At what distance along the z-axis is the electric field strength a maximum?

views

Domanda del libro di testo

The permanent electric dipole moment of the water molecule (H₂O) is $6.2 \times 10^{- 30}$ Cm. What is the maximum possible torque on a water molecule in a $5.0 \times 10^{8}$ N/C electric field?

2672

views