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Gauss' Law and charges

  1. Feb 2, 2006 #1
    A charge of 2pC is uniformly distributed throughout the volume between concentric spherical surfaces having radii of 1.3 cm and 3.3 cm. What is the magnitude of the electric field 1.8 cm from the center of these surfaces? Answer in units of N/C.

    I used the equation [tex] \Phi = E*4\pi r^2 [/tex] and then the equation [tex] \Phi = q_e_n_c /E_0 [/tex] and set them equal to each other. I used the formula q_e_n_c = charge/volume * volume to get [tex] Q/(4/3)\pi R^3 * (4/3)\pi r^3. [/tex] Simplifying gave me [tex] Qr^3/E_0 R^3 [/tex]. Plugging this in for q_e_n_c and solving for the E field gave me ,
    [tex] E= Qr/ E_0R^3*4\pi [/tex]
    Q= 2 x 10^-12 since the charge is uniform
    r =.018 m
    E_0 = 8.85 x 10^-12
    R= .033
    So 2 x 10^-12(.018)/8.85 x 10^-12 *(.033)^3 * 4pi = 9.02 N/C.... which isn't right.
    Can someone please help me? I have no idea what I'm doing. :confused:
     
  2. jcsd
  3. Feb 2, 2006 #2

    Astronuc

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    Staff: Mentor

    The electric field depends on the enclosed charge, and from the problem statement, it is the charge enclosed within the radius of 0.018 m, of the charge distribution between 0.013 m and 0.033 m. In other words, the charge enclosed in a shell of inner radius 0.013 m and outer radius 0.018 m.

    implies the charge 2 pC distributed over the entire spherical volume of radius 0.033 m.

    The 2 pC is distributed uniformly in a spherical shell 4/3[itex]\pi[/itex]((0.033 m)3 - (0.013 m)3).

    Then determine what charge is enclosed by the smaller spherical shell of inner radius 0.013 m and outer radius 0.018 m.
     
  4. Feb 2, 2006 #3
    Ok.. I'm not sure if I am doing this right but....
    [tex] q_e_n_c = 2 x 10^-12 * (.013)^3 / (.018)^3 [/tex]. Solving this would give me 7.53 x 10^-13 which is my Q. Then I plug this into my previous equation, [tex] Qr /E_0 R^3 4\pi [/tex]
    Where Q= 7.53 x 10^-13
    r= .013
    R = .018
    Is this the right way to do it?
     
  5. Feb 3, 2006 #4

    Astronuc

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    Staff: Mentor

    Think of how one finds the area between two circles, and by extension, the volume between two spherical surfaces.
     
  6. Feb 3, 2006 #5
    Ok so to find the area between the 2 circles, I would used [tex] 4\piR^3 - 4\pi r^2 [/tex] since they're spheres. Then I would multiply this answer by E for the equation for flux. I'm a little confused about the volume. Would I have [tex] Q/ (4/3)\pi (R^3 - r^3) * 4/3 \pi r^3 [/tex] ? I understand that I need to find the difference in volume between the 2 spheres, but I'm a little unsure of where it would go. Thanks
     
  7. Feb 3, 2006 #6

    Astronuc

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    Staff: Mentor

    Please be careful.

    The area between two circles is simply the difference in areas,

    A2 - A1 = [itex]\pi\,{r_2}^2\,-\,\pi\,{r_1}^2[/itex] = [itex]\pi\,({r_2}^2\,-\,{r_1}^2 )[/itex], with obviously r2 > r1

    and then likewise, the volume between two spheres is simply the difference in volumes,

    V2 - V1 = [itex]4/3\,\pi\,{r_2}^3\,-\,4/3\,\pi\,{r_1}^3[/itex] = [itex]4/3\,\pi\,({r_2}^3\,-\,{r_1}^3 )[/itex], again obviously with r2 > r1
     
  8. Feb 3, 2006 #7
    I'm sorry, I typed my previous post wrong. Anyway, I tried starting over using the help you gave.
    [tex]\Phi = E*A [/tex]
    [tex] \Phi= E * \pi (R^2 - r^2 [/tex]
    where R= .018 and r= .013
    So [tex] \Phi= 1.55 x 10^-4 \pi * E [/tex]
    Then [tex] \Phi= q_e_n_c / E_0 [/tex]
    [tex] Q_e_n_c = charge/volume * volume [/tex]
    [tex] Q_e_n_c= 2 x 10^-12 / (4/3)\pi (.033^3 -.013^3) * (4/3)\pi (.018^3 -.013^3) [/tex]
    So [tex] Q_e_n_c = 2.16 x 10^-13 [/tex]
    Then [tex] 1.55 x 10^-4 \pi *E = 2.16 x 10^-13/ 8.85 x 10^-12 [/tex]
    Solving for E gave me 50.1.
    Am I doing this right now?
     
  9. Feb 4, 2006 #8

    Astronuc

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    Staff: Mentor

    [tex] Q_{enc}= 2 x 10^-12 / (4/3)\pi (.033^3 -.013^3) * (4/3)\pi (.018^3 -.013^3) [/tex] is correct,

    however, I would write it

    [tex]Q\,\frac{0.018^3-0.013^3}{0.033^3-0.013^3}[/tex]

    in order to show clearly which is the numerator and which is the denominator. :smile:
     
  10. Feb 4, 2006 #9
    Is the first part of my flux wrong? It doesn't matter how I write the charge, I'm still getting 50.1 as my answer and it's wrong. I don't know how it can we wrong now!
     
  11. Feb 7, 2006 #10
    Can someone please tell me what I'm doing wrong?
     
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