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Homework Help: Electric Potential and Coulomb's Law Questions

  1. Feb 9, 2010 #1
    Any answers would be great, thanks!

    1. The problem statement, all variables and given/known data
    1. A charge Q is uniformly spread along the x-axis from x=0 to x=l. Find the electric potential function at points on the x-axis for x>l.

    2. An electron moves from one point to another where the second point has a larger value of the electric potential by 5 volts. If the initial velocity was zero, how fast will the electron be going at the second point? (The mass of the electron is 9.11E-31 kg)

    3. Suppose an electron were moving in a straight line towards a fixed nucleus which had a positive charge 8 times the size of the charge on an electron. If the electron started at infinity with essentially zero velocity, how fast would it be moving when it was 50E-11 m from the nucleus?

    2. Relevant equations

    Coulomb's Law = [tex]\alpha[/tex]*[q1*q2]/r^2
    where [tex]\alpha[/tex]=1/[4*[tex]\pi[/tex]*[tex]\epsilon[/tex][tex]_{}0[/tex]

    U=1/[4*pi*epsilon-sub0]*[q*q sub0]/r

    V=U/[q sub0]

    3. The attempt at a solution

    1. dV=1/[4*pi*epsilon]*[(Q/l)*dx]/[x-l]
    I don't know what to do with this one, the answer has a natural log (ln) in it.

    2. KE(initial) + U(initial) = KE(final) + U(final)
    0+1/[4*pi*epsilon]*q/r = 1/2 * m * v(final)^2 + 1/[4*pi*epsilon]*q/r
    Not sure what to do here either, seem to have three unknowns
    Answer = 1.33E6 m/s

    3. KE(initial) + U(initial) = KE(final) + U(final)
    0=1/2*9.109E-31(mass of an electron)*v(final)^2+9E9*[8*(1.6E-19)^2]/[50E-11]
    I don't understand this one either

    Thanks for all the help!
    Last edited: Feb 9, 2010
  2. jcsd
  3. Feb 9, 2010 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Let s be the position on the x axis at which the potential is measured. Write out the expression for potential due to the charge element Qdx/l at a position x where [itex]0 \le x \le l[/itex]

    How much kinetic energy does the electron acquire in moving to the second point? (Hint: for a negative charge, the second position is at a lower potential energy per unit negative charge by 5 joules/coulomb).

    What is the potential energy of the electron at infinity? What is it at the given distance? Where does that potential energy go?

  4. Feb 10, 2010 #3
    I'm still not getting question 4...
  5. Feb 10, 2010 #4
    There is no question 4.
  6. Feb 10, 2010 #5
    sorry it's 4 in my book, question 1
  7. Feb 10, 2010 #6
    The electric potential at point P, along the line, is kq/r d is the distance from the end of the rod to the point.
    [tex] dV = \frac{k}{x}dq [/tex]
    [tex] dq = \lambda dx = Q/L dx [/tex]
    [tex] dV = \frac{kQ}{Lx}dx [/tex]
    Then integrate from d to L + d.
    Last edited: Feb 11, 2010
  8. Feb 10, 2010 #7
    Ok thanks! Got it!
    I had:
    int[k*[(Q/L)dx]/(a-x)] from 0,L

    where L is the distance of where the charge is spread

    Thanks a lot!
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