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Physics Problem Please Help

  1. Jul 29, 2003 #1
    Heres The Problem:

    The electric field everywhere on the surface of a thin spherical shell of radius 0.750 m is measured to be equal to 890 N/C and points radially toward the center of the sphere.

    (a) What is the net charge within the sphere's surface?

    (b) What can you conclude about the nature and distribution of the charge inside the spherical shell?

    Here is where I am at:

    I know this has something to do with Gauss's Law

    r = .750 m

    E = 890 N/C

    Ke = 8.99 * 10^9 N*m^2/C^2

    Ke = 1 / 4 *(pi)*(permittivity of free space)

    (permittivity of free space) = 1 / 4*(pi)*(Ke)

    *is the Electric Flux the Net charge within the sphere's surface ?

    *q = E*(r^2)/Ke Is this correct?
    Am I going the right way on this?I Need To Solve This Algebraically
  2. jcsd
  3. Jul 29, 2003 #2
    I would approach this problem as a boundary value problem, not a Gauss' Law. If your teacher has covered boundary conditions, then you know there is one for the tangential E field and one for the normal E field. Which case do you have with this problem?

    The charge per unit area on the sphere is related to one of the boundary conditions.

    If the sphere is isolated, the total charge on it is constant. You know there can be no charge inside the metal of the sphere so that leaves only the inside and outside surfaces for charge.

    That should help some.
  4. Jul 30, 2003 #3
    I have absolutely no idea what a boundary problem is. We just started covering electricity. All I know is Gauss's Law and the Superposition Principle.
  5. Jul 30, 2003 #4
    I know what section of physics you are studying and you should be using Gauss's Law to solve this problem. This is a Gaussian surface problem.

    You should have available to you a table of typical electric field calculations using Gauss's Law. And within it you should find:

    Charge Distribution (description):
    Thin spherical shell of radius R and total charge of Q

    Electric Field:
    ke(Q/r^2) where r>R
    0 where r<R

    a) The electric field inside the spherical shell is zero. This follows from Gauss's law applied to a spherical surface of radius r<R (shown above). Since the net charge inside the surface is zero, and because of the spherical symmetry of the charge distribution, application of Gauss's law shows that E = 0 in the region r<R.

    b) See above.

    Now if you were looking for the electric field outside the shell, then you would be using the same law, but the E = ke(Q/r^2) equation will provide the answer. Just plug in and solve.
  6. Jul 30, 2003 #5
    I don't understand the whole r < R thing.
    How do I know that .750 < r? For that matter,How do I know that r < R or r > R in any case? All that was given to me was that r = .750 m. I read what you said in the book, but the book just doesn't elaborate. The whole concept just completely throws me off. I need someone to explain the r < R concept to me.
  7. Jul 30, 2003 #6
    Alright, I should've explained the variables better since I know they change between different text authors. So, here is a brief explanation of the variables I used:

    R = radius of the gaussian surface (in this case the sphere has an R of 0.750m.

    r = the radius of where you are evaluating the electric field. In your case, r < 0.750m.

    That's about it.

    What else is proving difficult for you regarding Gauss's laws?
  8. Jul 30, 2003 #7


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    ??? This assumes there in no net charge inside the surface to get the result that the electric field inside the shell is 0. The original question ASKED FOR the net charge inside the shell, not the field strength.
  9. Jul 30, 2003 #8
    If that is the case, then I am now even more confused than I originally was. Please clarify what you mean and what direction I would need to go in to fix the problem. Thank-you.
  10. Jul 30, 2003 #9
    Is the net charge Q ??? How would I get that???????
  11. Jul 30, 2003 #10
    And I gave the correct answer. With an adequate explanation (I thought). <shrug>

    Unless otherwise stated by the problem, there are NO charges (Q) inside the sphere, correct? To further explain my reasoning:

    Gauss's law deals with describing the relationship between the net electric flux through a closed surface (gaussian surface) and the charge enclosed by the surface.

    Gauss's law states that "the net electric flux through any closed surface is equal to the net charge inside the surface divided by the permitivity of free space." Therefore, a point charge located outside a closed surface will provide no net charge inside the surface because (typically, and provided there is no charge within the surface) the number of field lines entering the surface equals the number leaving the surface. Therefore, we can conclude that the net electric flux through a closed surface that surrounds no charge is zero.

    So, from symmetry: There is NO NET CHARGE within the spherical shell. The charge is on the shell of the sphere not from within the sphere and therefore we can treat this situation as a sphere (with no charge inside) surrounded by a uniform charge with field lines passing through the sphere. Thus creating a situation similar to that which I described in the previous paragraph.

    And since E = 0, then Qnet inside= 0.
  12. Jul 30, 2003 #11
    My question right now is How can one mathematically coclude that E or Qnet equals zero. I mean, somebody had to do it. I would like to know how. Is there a formula? I am sure that the book (or my teacher) would want to se some kind of mathematically-based solution, unless of course there isn't one. That would be assuming that r < R and E = 0. Again, I must ask, How do I know that R = .750 m and r < .750? How was that conclusion made? How would Qnet be computed? I really would like to understand this.
  13. Jul 30, 2003 #12

    Tom Mattson

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    I'm going to start from the beginning.

    First, use Gauss' law with the (spherical) Gaussian surface exactly at the outer wall of the shell. Gauss' law reads:


    You were told that the electric field over that surface is of constant magnitude, and you were also told that it points radially inward. That means that E and dS are antiparallel everywhere, so E.dS=EdS. That makes the integral easy, and that is in fact why you should use Gauss' law for this problem. With that, you can find the total charge on the shell.

    Second, as to the nature of the charge, there are two things to mention.

    1.) You know that the field lines are uniformly directed along radials. That tells you something about the charge distribution.

    2.) You know that the field lines are directed inwards. That tells you the sign of the charge (although you will already have figured that out in part a!).

    That's all there is to it.
  14. Jul 30, 2003 #13

    You know that R (radius of the spherical shell) = 0.750m because the problem stated such.

    You know that r (raduis of the field that you are evaluating) is less than R (=0.750m) because the problem statement asked for what the net charge is inside the spherical shell. Which means that r < R!

    Your physics book should have that table that I was talking about that shows the different solutions/equations to use for different scenarios.

    Your physics book should definitely have a section devoted to the different Gauss's law applications. As well as everything I just discussed in my last post.

    Which physics book are you studying from anyways? Everything that I posted can be found in "Physics for scientists and engineers" Vol. 2 fourth edition by Serway. That is the book that I studied from and the book that I refered all the students that I tutored to check out because it is excellent (see my threads in the Text book reviews forums).
  15. Jul 30, 2003 #14
    Tom, I need to be able to solve the problem using Algebra. I cannot use Calculus. Do you know of a way to do that???

    I am using College Physics, By Serway and Faughn.
  16. Jul 30, 2003 #15

    Tom Mattson

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    Yes, you can use a summation sign instead of an integral.

    Howsabout you state Gauss' law exactly as it appears in your book. Like they say in Jerry McGuire, "Help me help you".
  17. Jul 30, 2003 #16
    In my book Gauss's Law states that:

    (flux) = Q / e(o)

    Magnitude of electric field

    E = Ke |q| / r^2

    The electric flux through the surface:

    (flux) = EA = ke q / r^2 (4(pi)r^2)

    e(o) = 1 / 4(pi)Ke = 8.85 x 10 ^-12 = permittivity of free space

    the electric flux through the closed spherical surface that surrounds the charge q can be expressed as :

    (flux) = 4(pi)(ke)(q) = q / e(o)

    for Gauss's law they use a capital Q and for the other formulas the use a small Q so I am assuming the Q means the net charge as a whole, otherwise, that is how the formulas appear in my book.
    In the book they use the symbols for electric Flux and all but I just typed the word in parenthesis. I'm Sure you know what I mean. Thank-You For your help, I appreciate it.
  18. Jul 30, 2003 #17

    Tom Mattson

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    OK, so they give you the result of the integration. Now you can use that to calculate q. Note that flux into the Gaussian surface is negative.
  19. Jul 30, 2003 #18
    so, is the flux = 890 N/C

    and , would I use that to solve for q? Or, Am I going the wrong way?
  20. Jul 31, 2003 #19

    Tom Mattson

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    I don't have a calculator so I don't know about the magnitude, but the units are wrong. They should be Nm2/C

    Gauss' law says that Flux=q/&epsilon;0, so you are going the right way.
  21. Jul 31, 2003 #20
    My question is:
    Does the formula go like this:
    890 N/C = q / e(o) ?

    That was what I meant. Thank-you for your help.
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