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Guass's Law

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data


    Consider the electric field created by a very long charged line of negative linear charge density -2.50 nC/m.

    A small positive point charge of 8 mC moves from a distance of 9 cm to a distance of 17 cm.

    How much work is done by the electric field?

    Hint: The electric field for a long charged line is:

    The Equation is E line = 1/(4piE0)(2lamda/r)


    Express the result in the unit mJ and to three significant figures.
    2. Relevant equations


    The Equation is E line = 1/(4piE0)(2lamda/r)

    3. The attempt at a solution

    Is this hint equation provided equivalent to Gauss' Law: (2*pi*R*L*E = L*q/eo)

    where eo is the coulomb's law constant "epsilon-zero" = 8.89^10^-12 N*m^2/C^2
    and q is the charge per unit length

    Therefore,

    E(R) = q/(2 pi eo R)
    Multiply that by Q = 8 mC and integrate with dR from R = 0.09 to 0.17 m

    Also I am confused how to integrate this
     
  2. jcsd
  3. Feb 13, 2010 #2
    [tex]E = \frac{\lambda}{2 \pi \epsilon_o r}[/tex]
    [tex] W = \int\limits_{R_1}^{R_2} F_e\, dr[/tex]
    [tex] W = \int\limits_{R_1}^{R_2} qE\, dr[/tex]
    [tex] W = \int\limits_{R_1}^{R_2}\frac{q\lambda}{2 \pi \epsilon_o r}\, dr[/tex]
    Everything is constant except for the r.
    [tex] W = \frac{q\lambda}{2 \pi \epsilon_o }\int\limits_{R_1}^{R_2}\frac{1}{r}\, dr[/tex]
    [tex]W = \frac{q\lambda}{2 \pi \epsilon_o} ln(r)|\limits_{R_1}^{R_2}[/tex]
     
  4. Feb 14, 2010 #3
    Thanks so much,
    Do you know what the wavelength is
     
  5. Feb 14, 2010 #4
    [tex]\lambda[/tex] is linear charge density
     
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