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2 Problems regarding Gauss's Law

  1. Sep 8, 2005 #1
    Hey everyone, I have two problems that deal with Gauss's law. The first one deals with Electric Fields.

    Charge is distributed on a long, straight rod with uniform density 6.5 x 10^-8 C/m. Compare the magnitude of the field 1 cm from the rod to the field 1 cm from a point charge q = 6.5 x 10^-8.

    The answer in the back of the book is: E(rod)/E(point)=.02=2%

    I was thinking that the Electric fields would be the same since the charge is the same, but then I realized that the charge is spread out on the rod. How would I find the Field for the rod to solve the problem? I'm assuming that for the point charge it's just kq/r^2 = [(8.99 x 10^9)(6.5 x 10^-8)]/(.01^2).

    The second problem is regarding motion.

    Consider a solid sphere of radius R with a charge Q distributed uniformly. Suppose that a point charge q of mass m, with a sign opposite to that of Q, is free to move within a solid sphere. Charge q is placed at rest on the surface of the solid sphere and released. Describe the subsequent motion. In particular, what is the period of the motion, and what is the total energy of the point charge? [Hint: recall the properties of the motion for which the force varies linearly with the distance from a fixed point and is a restoring force.]

    If anyone could help me with either of these two problems, I would greatly appreciate it. Thank you!
  2. jcsd
  3. Sep 8, 2005 #2


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    Use guass law for the rod taking a coaxial cylinder of radius 1 cm as Guassian surface.
  4. Sep 8, 2005 #3
    Thanks mukundpa!
    You were right...the E(rod) is equal to (lamda)/[2(pi)(Epsilon)r]

    I plugged in all the coefficients and it worked perfectly =)

    Can anyone help me with the second problem?
  5. Sep 8, 2005 #4


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    For the second problem I think the charge distribution is volume charge distribution.
    The charge density rho = Q/(4*PI*R^3/3)

    Consider a spherical Guassion surface of radius x the charge within the Guassian surface will be rho*(4*PI*x^3/3) and hence fiels at distance x from the center will be
    E(x) =rho*(4*PI*x^3/3)/eo*(4*Pi*x^2) = Q*x/(e0*4*Pi*R^3) which is proprtional to x

    Now find the force actiong on the - ve charge and solve for simple harmonic motion.
  6. Sep 8, 2005 #5
    So taking E = Q*x/(e0*4*Pi*R^3), I substitute that into E = F/q?

    Therefore, F = Eq = Q*q*r/(e0*4*Pi*R^3)

    Using F how would I obtain the total energy of the point charge? Period of motion?
    Wouldn't the total energy be PE + KE?
    I'm really not that good at Physics, can you explain the concept for me?
  7. Sep 8, 2005 #6


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    First of all the charge q is nigative so the force will be negetive ( atteractive) and towards the center of the sphere.

    Secondaly q is released from the surface hence the amplitude is R, the radius of the sphere.

    Thirdly if F = - Kx the time period is 2Pi*sq.rt.(m/K)

    Total energy of the charge is only potential at the surface and remains conserved.
  8. Sep 8, 2005 #7
    Thanks! I get it now.
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