Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

JenniferSmart1

Gauss's Law

  • one year ago
  • one year ago

  • This Question is Closed
  1. darkb0nebeauty
    Best Response
    You've already chosen the best response.
    Medals 0

    What about it, can you please copy and past the question...please, or i can't help.

    • one year ago
  2. darkb0nebeauty
    Best Response
    You've already chosen the best response.
    Medals 0

    call me when you are ready..

    • one year ago
  3. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1359922050546:dw|

    • one year ago
  4. darkb0nebeauty
    Best Response
    You've already chosen the best response.
    Medals 0

    What is the question

    • one year ago
  5. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    My goodness have some patience...You can obviously see my typing... Find the electric field everywhere for a uniformly charged solid sphere that has a radius R and a total charge Q that is uniformly distributed throughout the volume of the sphere that has a charge density \[\rho=\frac QV\], where \[V=\frac 43 \pi R^3\] is the volume of the sphere. It's an example in my book. So far I understand that I need to make shell around the charged solid, and use that find the \[\phi_{net}\] and then we would find the Charge because \[Q=\phi_{net}\epsilon_0\] From that charge we find the electric field..... Here is my question What do they mean by \[\textrm{For r}\ge R, Q_{inside}=Q\] \[\textrm{For r}\le R, Q_{inside}=\rho V'\]

    • one year ago
  6. Jemurray3
    Best Response
    You've already chosen the best response.
    Medals 1

    They mean that once your shell goes beyond the radius of the charged sphere, the amount of charge inside is constant (obviously, just the total charge of the sphere). But when you're inside, the charge inside your shell is less than that.

    • one year ago
  7. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1359923152556:dw| can you show me where exactly? I'm am a visual learner.

    • one year ago
  8. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    "....once your shell goes beyond the radius of the charged sphere, the amount of charge inside is constant..." " But when you're inside, the charge inside your shell is less than that."

    • one year ago
  9. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1359923417543:dw| ?

    • one year ago
  10. Jemurray3
    Best Response
    You've already chosen the best response.
    Medals 1

    Just a moment...

    • one year ago
  11. Jemurray3
    Best Response
    You've already chosen the best response.
    Medals 1

    If the shell is INSIDE the sphere, then the charge enclosed is Q times the fraction of the sphere contained in the shell. If the shell is larger than the sphere, though, the total charge inside is just Q.

    • one year ago
  12. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    Since I made my shell outside the sphere....My charge inside is just Q, correct?

    • one year ago
  13. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    nope, I need to consider both, right?

    • one year ago
  14. JamesJ
    Best Response
    You've already chosen the best response.
    Medals 0

    Actually, insides the sphere the net charge should be zero, because all of the charge lies on the surface; i.e., \( \rho = 0 \). To see all of this explained, including this case, you might enjoy watching this excellent lecture: http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-3-electric-flux-and-gausss-law/

    • one year ago
  15. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    @JamesJ The specifically said "a uniformly charged solid sphere"

    • one year ago
  16. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    *they

    • one year ago
  17. JamesJ
    Best Response
    You've already chosen the best response.
    Medals 0

    ok. So the sphere is not a conductor.

    • one year ago
  18. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    I doubt it. What do you think @Jemurray3 ?

    • one year ago
  19. Jemurray3
    Best Response
    You've already chosen the best response.
    Medals 1

    No, it's not a conductor

    • one year ago
  20. JamesJ
    Best Response
    You've already chosen the best response.
    Medals 0

    So now... What do they mean by For r≥R,Qinside=Q For r≤R,Qinside=ρV′ What is meant is that when we are outside the sphere, the charge is the entire charge of the sphere Q. However, once you draw a Gaussian surface inside the sphere, you just want the charge within that surface, which is just the volume of that surface time the charge density. In other words, the charge outside of the surface is irrelevant.

    • one year ago
  21. JamesJ
    Best Response
    You've already chosen the best response.
    Medals 0

    I would still watch the lecture. He does still give you a lot of intuition for this sort of problem.

    • one year ago
  22. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    When we say Qinside, we mean the Q inside of the charged sphere, NOT the sphere that I have drawn around it, correct?

    • one year ago
  23. Jemurray3
    Best Response
    You've already chosen the best response.
    Medals 1

    Qinside means the charge inside the shell that you've drawn.

    • one year ago
  24. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    I think I get it now: The charge anywhere inside of the charged solid sphere is the density times the newly found volume (which depends on the size of r) The charge anywhere outside of that charged solid sphere, but still within the shell I have drawn is just Q.

    • one year ago
  25. Jemurray3
    Best Response
    You've already chosen the best response.
    Medals 1

    Yep

    • one year ago
  26. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    2 hours later.....I finally get it.

    • one year ago
  27. JenniferSmart1
    Best Response
    You've already chosen the best response.
    Medals 0

    awesome! Thanks guys

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.