1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Gauss' Law Question

  1. Jul 9, 2006 #1
    How would you solve this:

    A small charge of 443 C is at the center of a 7.97 cm radius ball. How much flux passes through the ball's surface?

    The answer is 4.922 E-8 N.m2/C

    I don't know how to get this answer. Please explain. Thank you!!!
  2. jcsd
  3. Jul 9, 2006 #2


    User Avatar
    Homework Helper
    Gold Member

    You wrote the title as "Gauss Law Question". Do you know what Gauss' law says? Can you apply it here?
  4. Jul 9, 2006 #3
    I know it is Gauss' Law because that is what the website I got it from said. I have looked everywhere and applied the Gauss' Law how I thought I should apply it but I keep getting the wrong answer and have no idea how they got that answer. So that is why I need someone to explain it for me if they know.
  5. Jul 9, 2006 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Perhaps if you showed your working we could point out where you have gone wrong...
  6. Jul 9, 2006 #5
    Well, I am confused as to what Gauss Equation to use... I have tried q=Eo*flux and then I tried flux=q/4pi*k and then I tried flux=k*q/r^2... none of them turn out correct.

    What is the equation to use for a point charge in a sphere?
  7. Jul 9, 2006 #6
    I keep finding that flux= q/Eo and so I put 443 in for q and 8.85*10^-12 for Eo and get 5.006*10^13.
  8. Jul 9, 2006 #7
    Assuming that you are familiar with calculus, you can have a look at http://en.wikipedia.org/wiki/Gauss_law. The flux is indeed [itex]q/\epsilon_{0}[/itex] in your case :smile: Do you have a doubt about the flux being independent of [itex]R[/itex]? If so, the wiki article/a reading from any standard textbook should clear it.
    Last edited: Jul 9, 2006
  9. Jul 9, 2006 #8


    User Avatar
    Homework Helper
    Gold Member

    443 C, is an awful lot of charge. Are you sure you got the numbers right from the question?

    You calculated the flux as q_enc/Eo, and that should work.
  10. Jul 9, 2006 #9
    I copied and pasted it so I know that is right. And I keep getting the wrong answer for ALL of the examples... which are all similar to that one.
  11. Jul 9, 2006 #10
    Another example on the webpage is:

    A small charge is at the center of a 9.77 cm radius ball. If 8.322 E-8 N.m2/C passes through the ball's surface, how much charge is at the center?

    and the answer is 749C.

    I don't get this one either. I am assuming the use the same equation.
  12. Jul 9, 2006 #11
  13. Jul 9, 2006 #12


    User Avatar
    Homework Helper
    Gold Member

    Ok, Gauss' law is

    [tex] \int \vec{E}.d\vec{a} = \frac{q_{enc}}{\epsilon_0} [/tex]

    That's the equation you're going to use for these problems.
    Now, note that
    (i) The value of [itex]\int \vec{E}.d\vec{a}[/itex] is the flux passing through the surface.
    (ii) Using Gauss' law, it follows from the previous step that [itex]\frac{q_{enc}}{\epsilon_0}[/itex] is also the flux passing through the surface.
    So, you've got 2 ways to calculate the flux.

    Why have they given the radius? Because you can calculate the area of the ball. Notice that, when a charge is at the center of a sphere, the integral [itex]\int \vec{E}.d\vec{a}[/itex] can be simply reduced to E times A. Can you figure out why, and solve the problems from here?
  14. Jul 9, 2006 #13
    no i have no idea what to do... i need step by step. and everything i read said you dont need the radius.
  15. Jul 9, 2006 #14


    User Avatar
    Homework Helper

    I think the problem is that whoever calculated the answers on that site is incorrectly using the Coloumb force constant {1/(4*Pi*epsilon_0)} instead of the permittivity of free space (epsilon_0) in their calculations. For instance, they ask this question:

    "A small charge of 45 C is at the center of a ball. If 5 E-9 N.m2/C passes through the ball's surface, what is the radius of the ball (if it matters)? "

    Now, obviously for that question the answer is that the radius doesn't matter, but if you use their values for charge and the flux, and plug into Flux = charge/k, solving for k you get 9x10^9, which is 1/(4*Pi*epsilon_0), not epsilon_0.

    Also, if in the original question you asked, if you say Flux = 443/(9x10^9), you'll find you get their answer.

    So, you were doing everything correctly; it's the website that is wrong.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook