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Electric Fields HW Question

  1. Sep 4, 2013 #1
    1. The problem statement, all variables and given/known data

    An infinitely long, uniformly charged straight line has linear charge density λ1 coul/m. A straight rod of length 'b' lies in the plane of the straight line and perpendicular to it, with its enared end at distance 'a' from the line. The charge density on the rod varies with distance 'y', measured from the lower end, according to λ(on rod) = (λ2*b)/(y+a), where λ2 is a constant. Find the electrical force exerted on the rod by the charge on the infinite straight line, in the λ1, λ2, a, and b, and constants like ε0.


    2. Relevant equations

    Gauss's Law.


    3. The attempt at a solution

    I first treated the problem as if there was only a point P above the infinite line and applied Gauss's Law using a cylinder as my Gaussian surface. My answer was

    E = (λ2 * b) / [2pi*ε0*(a+b)^2]

    From my understanding, I now have to relate my first answer to a change in distance of the point P along the rod. Any help would be appreciated.

    Thank you in advance.
     
  2. jcsd
  3. Sep 4, 2013 #2
    I hope I understood your problem correct!

    It's correct to use Gauss law to find the field of the line, but I think that you made a mistake because your answer is independent of [itex]\displaystyle{\lambda _1}[/itex]. Also your answer is not general, because point P is not a random point, it is rod's upper end (its distance from the line is [itex]\displaystyle{a+b}[/itex]).

    Now let's see a way to solve the problem.
    Firstly use Gauss law to find the field [itex]\displaystyle{E(r)}[/itex] of the line at a random point which distance from the line is a variable [itex]\displaystyle{r}[/itex]. This [itex]\displaystyle{r}[/itex] must be the radius of your Gaussian cylinder.

    Then take an infinitesimally small part of the rod, with length [itex]\displaystyle{dy}[/itex], which distance from the lower end of the rod is [itex]\displaystyle{y}[/itex] (and consequently from the line is [itex]\displaystyle{r=y+a}[/itex]). This part is charged with charge [itex]\displaystyle{dq}[/itex], so using [itex]\displaystyle{E(r)}[/itex] we can compute the force [itex]\displaystyle{dF}[/itex] exerted on it. Note that you can also find [itex]\displaystyle{dq}[/itex] in terms of [itex]\displaystyle{y}[/itex] and [itex]\displaystyle{dy}[/itex] using rod's charge density.

    Finally you get an equation for [itex]\displaystyle{dF}[/itex] in terms of [itex]\displaystyle{y}[/itex], [itex]\displaystyle{dy}[/itex] and some constants. You can integrate this equation to find the total force on the rod.

    I didn't use vectors for the field or the force because I think that the direction is trivial at this problem (everything upwards).
     
  4. Sep 4, 2013 #3
    The charge of the infinitesimally small part of the rod with lenght dy is λ(rod)*dy. Which is the force that acts on it? since you have found the electrical field that the line produces, then the force that exerts on a charge which lies in distance s from it is: dF = E(s)*dq.
     
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