Ch 23: The Electric Field

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 23: The Electric Field

Problem 37

Chapter 23, Problem 37

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.

Verified step by step guidance

Step 1: Identify the charges and their positions relative to the dot. The charges are: -5.0 nC at the top-left corner, +10 nC at the bottom-left corner, and +10 nC at the bottom-right corner. The distances from the dot to each charge are: 4.0 cm horizontally to the -5.0 nC charge, 2.0 cm vertically to the +10 nC charge at the bottom-left, and 4.0 cm horizontally to the +10 nC charge at the bottom-right.

Step 2: Use Coulomb's law to calculate the electric field contribution from each charge. The formula for the electric field due to a point charge is:

E = \frac{k q}{r^{2}}

, where

k

is the Coulomb constant (

8.99 \times 10 9^{N \cdot m^{2} / C^{2}}

q

is the charge, and

r

is the distance from the charge to the dot.

Step 3: Break the electric field contributions into components. For each charge, determine the direction of the electric field at the dot (away from positive charges and toward negative charges). Use trigonometry to resolve the electric field vectors into x and y components. For example, the electric field due to the -5.0 nC charge will have components in the negative x-direction, while the fields due to the +10 nC charges will have components in both x and y directions.

Step 4: Sum the components of the electric field. Add the x-components of the electric fields from all three charges to find the total x-component. Similarly, add the y-components of the electric fields to find the total y-component. This gives the electric field in component form:

E = (E x_{total}, E y_{total})

Step 5: Calculate the magnitude and direction of the electric field. The magnitude is given by:

E = \sqrt{E^{x_{total} 2} + E^{y_{total} 2}}

. The direction (angle) is given by:

θ = \tan ⁻ 1 (\frac{E}{y_{total}})

. Specify whether the angle is measured clockwise (cw) or counterclockwise (ccw) from the positive x-axis.

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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 directed away from positive charges and toward negative charges. The strength of the electric field (E) can be calculated using Coulomb's law, which states that the electric field due to a point charge is proportional to the charge and inversely proportional to the square of the distance from the charge.

Vector Components

Vector components are the projections of a vector along the axes of a coordinate system, typically the x and y axes in two-dimensional space. To analyze electric fields, it is essential to break down the electric field vector into its components, allowing for easier calculations of net electric fields from multiple charges. The components can be calculated using trigonometric functions, where the x-component is found using cosine and the y-component using sine.

Magnitude and Direction

The magnitude of a vector is a measure of its size or strength, while the direction indicates where the vector points. In the context of electric fields, the magnitude can be calculated using the Pythagorean theorem from the vector components, and the direction can be expressed as an angle relative to a reference axis, such as the positive x-axis. This angle can be measured clockwise (cw) or counterclockwise (ccw), which is crucial for accurately describing the orientation of the electric field.

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