1. Not finding help here? Sign up for a free 30min 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!

The Electric Field of a Ball of Uniform Charge Density

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A solid ball of radius r1 has a uniform charge density ρ.
    a) What is the magnitude of the electric field E(r) at a distance r>r1 from the center of the ball?
    b) What is the magnitude of the electric field E(r) at a distance r<r1 from the center of the ball?
    Express your answers in terms of ρ, r1, r, and ε.

    3. The attempt at a solution

    a) i was thinking since the question asks to evaluate the electric field from outside the ball you can treat it as a point charge, E=Q/(4∏ε)r^2, and since the question states it wants the answer in terms of ρ, i used the formula ρ=Q/V and changed the Q to Vρ. after plugging in the equation for V → (4/3)∏r^3, things cancel and your left with E=(1/3)ρr/ε , but that is wrong.

    i also tried this problem from the starting equation ∫EdA=Q/ε and got the exact same answer of E=(1/3)ρr/ε
     
  2. jcsd
  3. Sep 15, 2012 #2
    ok...i managed to solve for the answer to part a, however i don't understand the answer. the answer is E(r)=[(1/3)ρ(r1)^3]/r^2, i don't see how the radius of the ball (r1) has anything to do with the answer. to my knowledge it should have only been dependent on the radius of the gaussian surface i drew, which was r.
     
  4. Sep 15, 2012 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The total charge on the ball depends upon the radius of the ball as well as the charge density.

    [itex]\displaystyle Q_{\text{Ball}}=\left(\frac{4}{3}\pi\,{r_1}^3 \right)\rho\ .[/itex]
     
  5. Sep 15, 2012 #4
    is this because the ball is a nonconductor? since it has a charge density i cannot treat it as a point charge?
     
  6. Sep 15, 2012 #5

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Well, how did you do the calculation to come up with the correct answer?

    For any location exterior to the ball, i.e. r > r1 you can treat the charge as if it were a point charge at the center of the sphere.

    [itex]\displaystyle E(r)=\frac{1}{4\pi\epsilon_0}\frac{Q_\text{Ball}}{r^2}[/itex]
    [itex]\displaystyle =\frac{1}{4\pi\epsilon_0}\left(\frac{4}{3}\pi\,{r_1}^3\,\rho\right)\frac{1}{r^2}[/itex]

    [itex]\displaystyle =\frac{{r_1}^3\rho}{3\epsilon_0 r^2}[/itex]​
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook