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Homework Help: Potential difference problems

  1. Jan 4, 2004 #1
    Problem 4.
    Given: k_e=8.98755*10^9 nm^2/C^2 and g=9.8 m/s^2.
    find the potential differnece between a point infinitely far away and a point 1.4 cm from a proton. Answer in units of V.
    Note: I don't know where to start!

    Problem 6.
    Given: k_e=8.98755*10^9 nm^2/C^2 and g=9.8 m/s^2.
    Four particles with charges of 7.7 C, 4C, 3.3 C AND -5C ARE PLACED AT THE CORNERS of a (2.1m*2.1m) square.
    Determine the potential difference between the center of the square and infinity. Answer in units of V.
    Note: what formula should I use?

    Problem 7.
    Given: k_e=8.98755*10^9 nm^2/C^2 and g=9.8 m/s^2.
    An electron that is initially 57 cm away from a proton is displaced to another point.
    If the change in the electrial potential energy as a result of this movement is 2.2*10^-28 J, what is the final distance between the electron and the proton? Answer in units of m.
    Note: I don't know where to start.
     
  2. jcsd
  3. Jan 5, 2004 #2
    Give the formula for Potential at a point due to a point charge
     
  4. Jan 5, 2004 #3

    HallsofIvy

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    Science Advisor

    If you are not given a formula for potential energy due to a point source, you can find it by integrating the force function.

    Set up a coordinates system with the proton at the origin. The force on a unit charge at distance r from the origin is
    F= k/r2 (and has nothing to do with g)

    The anti-derivative is -k/r + C. Taking C=0 gives potential 0 at infinity (which is standard and is what your problem asks. Finally, take r= 1.4 cm.

    For problem 6, do the same thing with each of the 4 charges, using the distance from each charge to the center of the square as r.

    For problem 7, find the integral from 57 cm to X cm, set it equal to the given potential change and solve for X.
     
  5. Jan 6, 2004 #4
    Problem 6.

    As you said,"do the same thing with each of the 4 charges, using the distance from each charge to the center of the square as r."
    So you would use a^2+b^2=c^2 to find what r is? In addition, since to find r it seems to me that if i divide the square into four section each coming from one of the vertices and ending at the center; there would be 45/45/90 triangles and to find r would it just be 2.1/2=1.05=r. Right?
     
  6. Jan 6, 2004 #5
    roblem 7.

    What does integral mean in "find the integral from 57 cm to X cm"?
     
  7. Jan 7, 2004 #6
    Problem 7 Answer!

    This is what I did:
    2.2*10^-28=(8.99*10^9)((1.6*10^-19)(-1.6*10^-19)/r)
    2.2*10^-28r=(8.99*10^9)((1.6*10^-19)(-1.6*10^-19)
    2.2*10^-28r=1.736174035*10^-28
    r=1.267154073 or 1.3m
    However, this answer was incorrect, what did I do wrong?
     
  8. Jan 7, 2004 #7
    Two point charges

    problem 5.
    k_e=8.98755*10^9Nm^2/C^@ and g=9.8m/s^2.
    Two point charges of magnitude 6.7c and -3.4c are separated by 37.1 cm.
    what is the potential difference between a point infinitely far away and a point midway between the charges? answer in V.
    Note: I would use potential difference formula where r= 0.371/2=.1855. Potential difference=k_c*(q/r)
     
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