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physics problem -electricity and mag

Discussion in 'MCAT Discussions' started by proline, Jul 23, 2006.

  1. proline

    proline Junior Member

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    hey guys,

    i need help with this physics problem, its a little more complicated than what you will see on the MCAT but I'm not really sure what to do:


    An infinitely long line of charge has a linear charge density of 8.00×10−12 C/m. A proton is a distance of 17.5 cm from the line and moving directly toward the line with a speed of 2700 m/s.

    How close does the proton get to the line of charge?



    Does the proton stop before hitting the line? That would make sense since the repulsive positive charge of the line would repel the proton, but i don't understand how to find information about the particle at its final position? how would you determine either the potential energy or potential at its final destination?

    thanks for taking the time to read my question...
     
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  3. ih8mcat

    ih8mcat Dynomite!

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    do you know what the answer is supposed to be? I tried manipulating conservation of energy formula as follows:

    (kqQ/r)initial + (.5mv^2)initial = (kqQ/r)final + (.5mv^2)final

    THe KE final is zero, the only variable left is "r final"...this is probably beyond the scope of the MCAT, i ended up getting something like .16 m, so the distance between the two would be .175m - .16m

    I'M NOT SURE THOUGH
     
  4. gridiron

    Moderator Emeritus

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    You can't use that equation because it is a line of charge, so a integral is needed to solve for the full effect of the electric potential.
     
  5. gridiron

    Moderator Emeritus

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    I'll try to take a stab at this.......you have line of charge with a uniform linear density. I will assume it is a thin nonconducting rod of length L and has a positive charge of uniform linear density lambda. If say the proton is a perpendicular distance d from the left end of the rod and consider a differential element dx of the rod then:

    dq = lambda dx​

    This element produces a electric potential dV at point P which, a drawing will help here--the file I have is too big, is a distance r= (x^2 + d^2)^0.5. Than the potential dV is:

    K dq/r = k lambda dx / (x^2 + d^2)^0.5)​

    You would integrate the above equation from 0 to L==the length of the rod and get:

    V = k lambda* ln[ (L + (L^2 +d^2)^0.5/ d]​

    Since the proton is moving toward this rod, than a external force is required to push the proton against the electric field set up by this uniform linear charge. You would integrate the differential electric field and then set it equal to the change in potential do find the final distance. In all, multiple integrals are involved, so you will not see this on the MCAT!
     
  6. grapeflavorsoda

    grapeflavorsoda Senior Member

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    ummm. the length of the rod is infinite. so integration's upper limit would be infinity and lower limit would be -(infinity). so whole thing is wrong.
    better way would be to find the function of electric field in terms of perpendicular distance from the line.
     
  7. grapeflavorsoda

    grapeflavorsoda Senior Member

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    i assume that you are taking calc. based intro physics?

    if then use the gauss's law of electric flux => charge enclosed is proportional to the enclosed surface flux surrounding the charge.

    since it is infinite line with constant charge, electric field will have the line symmetry. use cylinder-like gaussian surface to calculate the field.

    let L = linear charge density.
    p = permittivity constant.
    E = electric field.
    h = distance(orthogonal) away from the line.
    (pi) = constant pi
    d = length of the wire
    thus putting gauss's law in terms of mathematical symbols

    (Ld)/p = electrical flux through cylinder wrapping the infinite line.

    therefore (Ld)/p = 2(pi)hdE => E = L/(2p(pi)h)

    so electric field is a function of "h" only.

    do the line integral with E and the distance vector that is radiating from linear symmetry, where the center of the symmetry is taken as zero. and multiply the result by negative one; remember that potential energy is negative value of the line intergal of the conservative force vector and distance.

    when you integrate V = -[L/2p(pi)]ln(h)

    multiplying V with charge gives the potential energy at points away from the linear symmetry center. You can just calculate the kinetic energy of the moving proton and see at what value of d, potential energy is equal to kinetic energy.

    using this
     
  8. grapeflavorsoda

    grapeflavorsoda Senior Member

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    at the last line of the post "d" should be "h". habit of designating d with distance. unfortunately i chose h to be the distance.
     
  9. gridiron

    Moderator Emeritus

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    I was wrong the first time....but through PM I gave proline a different approach that would work.....yes I was wrong the first time in some of my assumptions
     

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