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Homework Help: Finding the Net Charge of the Earth

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data

    The Earth is surrounded by an electric field, pointing inward at every point. Assume a magnitude of E = 142N/C near the surface.
    a) What is the net charge on the Earth?
    b) How many excess electrons per square meter on the Earth's surface does this correspond to?

    E = 142 N/C
    Radius of Earth = 6.3781 km = 6,378,100 m

    2. Relevant equations

    Gauss's Law = EA

    3. The attempt at a solution

    So i know that i have to use Gauss's Law here. Which is simple the Electric Field, (given) and the area of the earth which is found be 4πr2

    Gauss's Law = EA
    = (142) * (4πr2)
    = (142) * (4π(6,378,100)2)
    = (142) * (5.11e+14)
    = 7.26e+16 C

    however. After attempting that, i found it wasn't the right answer. Where did i go wrong?

    i haven't attempted the second part since i need to know the first part. But i assume that to do it i would have to find the surface area of the earth and divide it by the number i found in part a, then divide that by 1.6e-19 to find the number of electrons?
  2. jcsd
  3. Oct 10, 2011 #2
    Nevermind. I solved the problem.
  4. Oct 10, 2011 #3
    If you don't mind, please post the solution and what brought you to it. It's always helpful for others who might have a similar question to be able to see how others having the same issue might successfully approach it.
  5. Oct 10, 2011 #4
    Gauss's Law = EA
    = (142) * (4πr2)
    = (142) * (4π(6,378,100)2)
    = (142) * (5.11e+14)
    = 7.26e+16 C

    θ = 180° for each ∆A => -1
    (Basically means it's negative since electric field and ∆A are in complete opposite directions So Electric Field then is just a negative)

    Gauss's Law = q / E0
    q = (-E * A * E0 )
    q = -E * E0 * Area
    q = -[142] * [8.85e-12] * [4π(6,378,100)2]
    q = [-7.26e+16 C] * [8.85e-12 C2 / Nm2]
    q = -6.425e+5

    Number of Electrons (Part b)

    We know the electric field of the earth = -6.425e+5
    That is -6.425e+5 / -1.6e-19 = 4.02e+24 Electrons
    The Earth is 5.11e+14 Square meters

    5.11e+14 / 4.02e+24
    = 7.86e9 Electrons per Square Meter
    Last edited: Oct 10, 2011
  6. Oct 10, 2011 #5
    Looks good. Thank you. For future reference, by the way, you can simply write Gauss's law as flux = closed integral of E dot dA = qencl. / epsilon0 (and using the TeX tags you can represent flux with the capital Phi using ' \Phi ' and lowercase epsilon w/ subscript 0 ("epsilon naught") as ' \epsilon_{0} ' -- case-sensitive for both of those, and the closed integral as ' \oint_{S} ' to show it's the closed integral over the surface).

    You can also simply rewrite it in ASCII if you're in a hurry as

    q_enc = eps_0 * (E . dA)

    where the boldface letters represent vector quantities.

    One of the great things about PF is that it has a wide range of easy-to-learn tools to let you express your mathematics clearly.

    But yeah, you got the answer right. At this point I'm basically just talking style: It's less confusing to the reader when you say "flux" or "Phi" equals this or that instead of saying "Gauss's Law" equals something.
  7. Oct 15, 2012 #6
    Also is it possible to show how your units cancel out to be Coulombs only for the net charge of the earth? It would help alot since Im trying to figure out how the units cancel. Thanks in advance.

    Also for your last part (5.11e14)/(4.02e24) =/= 7.89e9.
    (5.11e14)/(4.02e24) = 1.27e-10.

    I think you just switched the numbers around wrong but the answers are correct.
    Last edited: Oct 15, 2012
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