Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gaussian surface problem help needed

  1. Sep 1, 2006 #1
    I just realized that I accidentally didn't post this in the homework help section...if any mods can move this, I'd appreciate it. Sorry for the trouble!

    Hello. I need help solving the following problem:

    Consider a spherical Gaussian surface and three charges: q1 = 1.60 μC, q2 = -2.26 μC, and q3 = 3.71 μC. Find the electric flux through the Gaussian surface if it completely encloses the following charges.

    (a) only charges q1 and q2
    (b) only charges q2 and q3
    (c) all three charges
    (d) Suppose a fourth charge, Q, is added to the situation described in part (c). Find the sign and magnitude of Q required to give zero electric flux through the surface.

    I'm pretty sure that this problem involves the equation φ = q/ε0, but have no idea on how to incorporate the given values into this problem. Any help would be greatly appreciated. Thanks in advance!
    Last edited: Sep 1, 2006
  2. jcsd
  3. Sep 1, 2006 #2
    Maybe writing the equation this way would help: [tex]\Phi = \frac{Q_{enclosed}}{\epsilon_0}[/tex]
  4. Sep 1, 2006 #3
    Hmm...would you be able to give me an example of how to plug in my values into the equation, such as for part (a). I can try figuring out how to solve it for the rest if I get a sample with part (a). (There are no examples for this problem in both my notes and the textbook, so I have no clue on how to plug in everything.)
  5. Sep 1, 2006 #4
    Q_enclosed is nothing but the net charge enclosed within the surface. So, in case (a), what is the net charge enclosed?
  6. Sep 1, 2006 #5
    So for part (a) it would be 1.60μC - 2.26μC = -0.66μC? Shouldn't the answer be in Nm^2/C?
  7. Sep 1, 2006 #6
    Yes, and yes.
  8. Sep 1, 2006 #7
    Okay, so (-0.66 x 10^6 C)/(8.85 x 10^-12 C^2/N x m^2) = -7.74576 x 10^16 Nm^2/C is my answer for part (a)?
  9. Sep 1, 2006 #8
    The charge is given in terms of micro-coulombs, not Mega-, or million-coulombs. And it's ~ -7.4576.
  10. Sep 1, 2006 #9
    Okay, I changed my mistake of not putting the - in when converting form micro to standard. Is the final answer -74,576.27 x 10^16 Nm^2/C?
  11. Sep 1, 2006 #10
    No, it's -7.4576 x 104 Nm2/C.
  12. Sep 1, 2006 #11
    Hmm...what did I do wrong?
  13. Sep 1, 2006 #12
    Quite simple, but if were to do the same in a test, it would get you into trouble. ;) You didn't add the powers of ten correctly.
  14. Sep 1, 2006 #13
    LOL okay, thanks! So I will follow the same format for (b) and (c). Any ideas on how to approach (d)?
  15. Sep 2, 2006 #14
    (d): It just asks for another charge that would make the flux zero, i.e. net charge Q = 0, since [tex]\epsilon_0[/tex] is a constant.
  16. Sep 2, 2006 #15
    So is my answer -3.05μC! Thanks so much for your help! =]
    Last edited: Sep 2, 2006
  17. Sep 2, 2006 #16
    That's correct. Sorry for the late reply.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook