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Homework Help: Gauss' Law and electric fields

  1. Jul 21, 2008 #1
    I really just want to know if I'm doing this correctly, as I don't have access to the answer to check.

    1. The problem statement, all variables and given/known data

    A cylindrical shell of radius 7.00 cm and length 240 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 19.0 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. Find (a) the net charge on the shell and (b) the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.

    2. Relevant equations

    [tex]\phi=4\pi kQ[/tex]

    3. The attempt at a solution

    (a) Because 19.0 cm is relatively close to the shell given its length of 240 cm, I chose a cylinder with radius of 0.19 m as my Gaussian surface. The electric field vector is parallel to the normal vector of the cylindrical surface at all points, so

    [tex]\phi=EA=E2\pi (r-0.07 m)h=4\pi kQ \frac{h}{2.4 m}[/tex]

    [tex]Q=\frac{(2.4 m)E(r-0.07 m)}{2k}=\frac{(2.4 m)(3.60 * 10^{4} N/C)(0.12 m)}{2(9 * 10^{9} Nm^{2}/C^{2})}=5.76 * 10^{-7} C[/tex]

    (b) A point 4.00 cm from the axis is within the shell itself since it has a radius of 7.00 cm. If I were to choose a cylinder of radius 0.04 m for a Gaussian surface, would the electric field not be zero since it would contain no charge within?

    I'm really not sure if I'm modeling this correctly at all. Any help is appreciated.
    Last edited: Jul 21, 2008
  2. jcsd
  3. Jul 21, 2008 #2


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    Hi Tentothe,

    The area A is the area of the Gaussian surface, which is a cylinder with radius 19cm. Remember in this case we want the surface area of the curved part of the cylinder. Do you see what's not right in the above work?

    That sounds right to me.
    Last edited: Jul 21, 2008
  4. Jul 21, 2008 #3
    I'm guessing it's the radius that's throwing me here. I think I may have originally had it correct with:

    [tex]Q=\frac{(2.4 m)Er}{2k}=\frac{(2.4 m)(3.60 * 10^{4} N/C)(0.19 m)}{2(9 * 10^{9} Nm^{2}/C^{2})}=9.12 * 10^{-7} C[/tex]

    My thinking was that I'd made a mistake because the radius of the shell itself (7 cm) wasn't used anywhere in my answer, but it looks like that's really only there to help with part (b).
  5. Jul 21, 2008 #4


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    That looks right to me.
  6. Jul 21, 2008 #5
    Thanks for your help!
  7. Jul 21, 2008 #6


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    Sure, glad to help!
  8. Nov 18, 2010 #7
    please help me for b
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