Question

$\mathrm{A}-10 \mathrm{nC}$ charge is located at the origin. What are the strengths of the electric fields at the positions $(x, y)=(0.0 \mathrm{cm}, 5.0 \mathrm{cm}),(-5.0 \mathrm{cm},-5.0 \mathrm{cm}),$ and $(-5.0 \mathrm{cm}, 5.0 \mathrm{cm}) ?$ b. Draw a field diagram showing the electric field vectors at these points.

Name: Problem 27, Chapter 20: College Physics: A Strategic Approach
Uploaded: 2020-12-02T03:27:50-08:00
Duration: 5 min 28 s
Channel: Khoobchandra Agrawal
Description: A-10 nC charge is located at the origin. What are the strengths of the electric fields at the positions (x, y)=(0.0 cm, 5.0 cm),(-5.0 cm,-5.0 cm), and (-5.0 cm, 5.0 cm) ? b. Draw a field diagram showing the electric field vectors at these points.

   $\mathrm{A}-10 \mathrm{nC}$ charge is located at the origin.
What are the strengths of the electric fields at the positions $(x, y)=(0.0 \mathrm{cm}, 5.0 \mathrm{cm}),(-5.0 \mathrm{cm},-5.0 \mathrm{cm}),$ and
$(-5.0 \mathrm{cm}, 5.0 \mathrm{cm}) ?$
b. Draw a field diagram showing the electric field vectors at these points.

College Physics: A Strategic Approach

Randall D. Knight,… 4th Edition

Chapter 20, Problem 27

Instant Answer

Solved by Expert Khoobchandra Agrawal

12/02/2020

Step 1

Use E = k|q|/r^2 with k = 8.99×10^9 N·m^2/C^2 and q = −10.0 nC = −1.0×10^−8 C. Convert cm → m (5.0 cm = 0.05 m). - Point (0.0 cm, 5.0 cm) → (0, +0.05 m): r = 0.05 m, E = k|q|/r^2 = (8.99×10^9)(1.0×10^−8)/(0.05^2) = 3.596×10^4 N/C ≈ 3.60×10^4 N/C. Direction: q Show more…

Show all steps

Thanks for your feedback!

$\mathrm{A}-10 \mathrm{nC}$ charge is located at the origin. What are the strengths of the electric fields at the positions $(x, y)=(0.0 \mathrm{cm}, 5.0 \mathrm{cm}),(-5.0 \mathrm{cm},-5.0 \mathrm{cm}),$ and $(-5.0 \mathrm{cm}, 5.0 \mathrm{cm}) ?$ b. Draw a field diagram showing the electric field vectors at these points.

Play audio

Feedback

Khoobchandra Agrawal and 70 other subject Physics 102 Electricity and Magnetism educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Electric Field Due to a Point Charge

This concept involves understanding that a point charge creates an electric field in the surrounding space, whose strength and direction are given by Coulomb's law. The electric field at any point is calculated as E = k|q|/r², where k is Coulomb’s constant, q is the charge, and r is the distance from the charge. This inverse square relationship indicates that the field strength diminishes with the square of the distance from the charge.

Vector Representation of Electric Fields

Electric fields are vector fields, meaning that they have both magnitude and direction. When representing electric field vectors graphically, each vector is drawn with its tail at the point of interest and points in the direction in which a positive test charge would be accelerated. This involves decomposing the field values into components if needed, and visually interpreting the field’s behavior across space.

Please give Ace some feedback