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Electric Field and Charge Density

  1. May 4, 2008 #1
    1. The problem statement, all variables and given/known data
    A line of charge starts at [tex]x=+x_0[/tex] and extends to positive infinity. If the linear charge density is [tex]\lambda=\frac{\lambda_0x_0}{x}[/tex], determine the electric field at the origin.

    3. The attempt at a solution

    I'm really not sire which equations I would have to use to solve this...I am really just stuck at looking for a starting point to approach this.

    I was thinking maybe [tex]E=k_e\int\frac{dq}{r^2}[/tex] might be of some use. I'm not sure how to associate electric fields with charge density, and I'm not quite sure how the infinity plays into this either. Any help would be greatly appreciated, thanks so much.
     
  2. jcsd
  3. May 4, 2008 #2
    Any pointer would be greatly appreciated.
     
  4. May 4, 2008 #3
    That seems like the right equation. Gauss's law doesn't seem suitable so stick to that equation you have there. The linear charge density, L (landa) is q/x (q is charge, x is length), so the derivative of the linear charge density is dq/dx, which is equal to, in this case, -L0*x0/x^2 (the derivative of the expression on top). Then simply do a change of variable: dq = -L0*x0/x^2*dx and substitute that in for dq in the integral. The r^2 refers to the square of the distance the point at the origin is from each increment of length dx along the rod. If you think about it, whatever the value of x is, its going to be the same value as r, so r^2 is the same as x^2 in this case. From this point forward, you simply integrate.
     
  5. May 4, 2008 #4
    Thanks for the help. I'll try and do that to see how it works out.
     
  6. May 4, 2008 #5
    Alright. The domain of the integral is simply along the length of the rod.
     
  7. May 4, 2008 #6
    The charge is along a line that goes to infinity though...how would that work?
     
  8. May 4, 2008 #7

    alphysicist

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    Hi Gear300,

    I think you have a small error here. It's not that the derivative of lambda with respect to x is equal to dq/dx; lambda itself is equal to dq/dx. So you would get

    [tex]
    dq = \frac{\lambda_0 x_0}{x} dx
    [/tex]
     
  9. May 4, 2008 #8

    alphysicist

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    Hi XxBollWeevilx,

    Gear300 answered that in post #5; the endpoints of the line charge give the limits on the integral.
     
  10. May 5, 2008 #9
    Right, I get that now. The integral would be evaluated from +x to infinity. Could someone point me to a place, or perhaps explain to me briefly, how substituting an infinity into an expression would affect it? I know that if infinity is ever in the denominator then the expression would be 0. Are there any other important rules that might help me out?

    Again, thanks Gear and al.
     
  11. May 6, 2008 #10
    0 in the denominator reaches to infinite. Infinite over infinite is indeterminate, along with infinite over zero and zero over infinite.
     
  12. May 6, 2008 #11

    alphysicist

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    I don't think limits that gave zero/infinite would be indeterminate if it appeared in this solution; the limit would just be zero.
     
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