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: Electric current and resistance

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Two wires are connected to a conducting sphere of radius 7.75 cm, which is initially uncharged. One wire carries a current of 3.47 μA into the sphere, and another wire carries a current of 1.26 μA out of the sphere. How long does it take to to produce an electric potential of 5.00 V at a distance of 11.6 cm away from the center of the sphere?

    2. Relevant equations

    Csphere = 4 pi (8.85e-12) [(ra*rb)/(ra-rb)]

    C = Q/V



    Vsphere = KQ [-1/r]rarb ~(this doesn't look quite right but it's rb plugged in minus ra)

    3. The attempt at a solution

    I found that Csphere = -2.35e-11 and that when I plugged 5.00V into the Vsphere equation I came up with Q = 1.3e8.

    I'm just lost as to where to go from here. I can't seem to find anything linking together the pieces I have. I tried using the C=Q/V but I just kept going in circles.
  2. jcsd
  3. Feb 23, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You only have one conductor (a single sphere with a single radius), not a spherical shell with an inner and outer radius. So, doesn't the electric potential anywhere outside the sphere depend only on the total amount of charge and distance from centre of sphere [EDIT: as though it were a point charge]? In that case, Coulomb's law would be easy to apply.
    Last edited: Feb 23, 2010
  4. Feb 23, 2010 #3
    I guess I don't see how Coulomb's law applies. I don't have a point charge at all. If it's conducting, I have a uniform charge across the entire sphere and I don't even know what that charge is, just that so much current goes in and so much comes out.

    Plus doesn't coulomb's law determine force? I know V=[Force * (change in x)]/q, but how can that be applied to the time it takes to produce an electric potential?

    If I'm supposed to use Coulomb's Law ~ F = (9e9)Q/(11.6-7.75) How can I calculate Q to make this work? and how do I relate back to electric potential and time?
  5. Feb 23, 2010 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, no, but you have a distribution of electric charge that is spherically symmetric. Hence, by Gauss's law, the field outside the sphere at distance r is the same as the field that would be produced by a point charge of the same magnitude located at the centre of the sphere.

    You know the rate at which charge is accumulating onto the sphere. Therefore, if I give you any time interval since the current first started, you can tell me how much total charge has accumulated. The objective of this problem is to determine the time interval required to accumulate a specific amount of charge.

    Er, yeah, I was playing fast and loose with terminology. Coulomb's law gives you the electric field (or force, if you like). I meant that you can use the equation for the electric potential of a point charge:


    Now, due to the mathematical relationship between electric field and electric potential, this statement is logically equivalent to Coulomb's law (applied to a point charge).
  6. Feb 23, 2010 #5
    Okay so this is what I tried.

    3.47-1.26=2.21[tex]\mu[/tex]A ~ this is the rate at which the charge is building

    [tex]\Delta[/tex]V = E [tex]\Delta[/tex]x

    E = KQ/r2

    so V = KQ/r2 * r = KQ/r

    5 = [(9*109) Q] / 3.85 [tex]\rightarrow[/tex] Q = 2.14*109

    I = q/t

    2.21*10^-6 = 2.14*10^9/t [tex]\rightarrow[/tex] t = 9.68*10^14

    This is not the correct answer so I'm guessing I used the wrong equation or messed up my math somewhere, but I'm not seeing it. Sorry I keep coming back to this.
  7. Feb 23, 2010 #6
    I figured it out. I was using the wrong radius.

    Thanks for the help :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook