Question

The figure below shows a section of a very thin, very long, straight rod with a uniform charge per unit length of \lambda. Point O is a perpendicular distance $d$ from the rod. A spherical gaussian surface is centered at point O and has a radius R. (Use any variable or symbol stated above along with the following as necessary: \epsilon_0) (a) What is the electric flux through the spherical surface if $R < d$? \Phi_E = (b) What is the electric flux through the spherical surface if $R > d$? \Phi_E =

          The figure below shows a section of a very thin, very long, straight rod with a uniform charge per unit length of \lambda. Point O is a perpendicular distance $d$ from the rod. A spherical gaussian surface is centered at point O and has a radius R. (Use any variable or symbol stated above along with the following as necessary: \epsilon_0)

(a) What is the electric flux through the spherical surface if $R < d$?
\Phi_E = 

(b) What is the electric flux through the spherical surface if $R > d$?
\Phi_E =
        
Show more…
The figure below shows a section of a very thin, very long, straight rod with a uniform charge per unit length of λ. Point O is a perpendicular distance d from the rod. A spherical gaussian surface is centered at point O and has a radius R. (Use any variable or symbol stated above along with the following as necessary: ϵ0)

(a) What is the electric flux through the spherical surface if R < d?
= 

(b) What is the electric flux through the spherical surface if R > d?
=

Added by Chelsey S.

Close

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
Point O and has a radius R. (Use any variable or symbol stated above along with the following as necessary: \\epsilon _(0).)\ (a) What is the electric flux through the spherical surface if \\Phi _(E)=R>d?\ \\Phi _(E)=R\ \\Phi _(E)=\ (b) What is the electric flux through the spherical surface if R>d?\ \\Phi _(E)=\ Need Help? The figure below shows a section of a very thin, very long, straight rod with a uniform charge per unit length of 2. Point O is a perpendicular distance d from the rod. A spherical Gaussian surface is centered at point O and has a radius R. (Use any variable or symbol stated above along with the following as necessary:)\ (a) What is the electric flux through the spherical surface if R<\ DE\ (b) What is the electric flux through the spherical surface if R > d?\ DE\ Need Help? Read Submit Answer
Close icon
Play audio
Feedback
Powered by NumerAI
Danielle Fairburn Ivan Kochetkov
Jennifer Stoner verified

Krishna G and 81 other subject Physics 101 Mechanics 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

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.

*

Recommended Videos

-
two-concentric-imaginary-spherical-surfaces-of-radius-r-and-2r-respectively-surround-a-positive-point-charge-q-located-at-the-center-of-the-surfaces-when-compared-to-the-electric-flux-1-thro-98283

Two concentric imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge -Q located at the center of the surfaces. When compared to the electric flux ̄́1 through the surface of radius R, the electric flux ̄́2 through the surface of radius 2R is ̄́2 = ̄́1/4. When a cube is inscribed in a sphere of radius r, the length L of a side of the cube is L = 3r. If a positive point charge Q is placed at the center of the spherical surface, the ratio of the electric flux ̄́sphere at the spherical surface to the flux ̄́cube at the surface of the cube is 1/3.

Krishna G.

the-figure-below-shows-a-section-of-a-very-thin-very-long-straight-rod-with-a-uniform-charge-per-unit-length-of-a-point-0-is-a-perpendicular-distance-d-from-the-rod-a-spherical-gaussian-surf-23368

The figure below shows a section of a very thin, very long, straight rod with a uniform charge per unit length of λ. Point O is a perpendicular distance d from the rod. A spherical gaussian surface is centered at point O and has a radius R. (Use any variable or symbol stated above along with the following as necessary: ε₀.) a) (6 points) What is the electric flux through the spherical surface if R < d? b) (4 points) What is the electric flux through the spherical surface if R > d?

Penny R.

a-sphere-of-radius-r-surrounds-a-point-charge-q-located-at-its-center-a-show-that-the-electric-flux-86015

A sphere of radius R surrounds a point charge Q , located at its center. (a) Show that the electric flux through a circular cap of half-angle ) (Fig. P24.53) is $$\Phi_{E}=\frac{Q}{2 \epsilon_{0}}(1-\cos \theta)$$ What is the flux for $(\mathrm{b}) \theta=90^{\circ}$ and $(\mathrm{c}) \theta=180^{\circ} ?$

Eduard S.


*

Recommended Textbooks

-
University Physics with Modern Physics

University Physics with Modern Physics

Hugh D. Young 14th Edition
achievement 1,071 solutions
Physics: Principles with Applications

Physics: Principles with Applications

Douglas C. Giancoli 7th Edition
achievement 1,782 solutions
Fundamentals of Physics

Fundamentals of Physics

David Halliday, Robert Resnick , Jearl Walker 10th Edition
achievement 1,493 solutions

*

Transcript

-
00:01 In this problem it is given that two concentric imaginary spherical surface of radius are, radius is r and 2r, respectively surround a positive point charge minus q.
00:16 At the center of the surface and when compared to the electric flux 5 -1 through the surface of radius r when the electric falls 5 -2 through the surface of radius is 2 -through -the -surface of radius is 2 -r.
00:30 Have to find this so suppose there are two spheres like this this is the center and this is the radius are and this is minus theta so we know pi is equals to submission of q by submission of theta so pi 1 is equals to submission of q by submission of theta.
01:03 So this is equals to minus q by e theta...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Join the community

18,000,000+

Students on Numerade


Trusted by students at 8,000+ universities

Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever