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Force on Particle due to Sphere with Cylindrical Hole

  1. Jan 24, 2009 #1
    Note: This is my first time using LaTeX. Any formatting advice would be appreciated.

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

    Consider a homogeneous sphere of radius R_s made of a material that exerts a force obeying the inverse r-squared law (i.e. a particle of this material exerts a force on a test particle [tex]\vec{F}=\frac{K}{r^2}\hat{r}[/tex], where the vector r points from the particle to the test particle and K is a real number such that positive values are repulsive and negative values attractive). Next, consider a cylindrical hole of radius R_d that is removed from the sphere such that the axis of the cylinder passes through the center of the sphere. The hole passes completely through the sphere. Finally, consider a test particle arbitrarily placed on the axis of the cylinder a distance R from the center of the sphere. Determine the force felt by the test particle due to this configuration.

    2. Relevant equations



    3. The attempt at a solution

    Due to the cylindrical hole, I decided to construct the sphere by integrating cylindrical shells. I defined the x-axis to be the axis of the cylinder and the y-axis to be orthogonal to the x-axis. Considering two particles equidistant from the test particle and the x-axis, I found the force on the test particle to be [tex]\vec{F} = \frac{2 K r_{x}}{r^3}\hat{x}[/tex], with [tex]r=\sqrt{r_{x}^2+r_{y}^2}[/tex] and r_x and r_y being the x- and y-components of the distance from the particles to the test particle, respectively.

    Next, I revolved this about the x-axis to produce a ring of material of radius r_y. Now, [tex]\vec{F}=\frac{2 K r_{x}}{r^3}\hat{x} \pi r_{y} n[/tex], where [tex]n[/tex] is a number density term. I then note that the distance from the center of the sphere to the center of the ring is [tex]r_{d}=\sqrt{R_{s}^2+r_{y}^2}[/tex] and that r_x = R - r_d. I can then rewrite the force caused by the ring as [tex]\vec{F}=\frac{2 \pi K n (R-r_{d})\sqrt{R_{s}^2-r_{d}^2}}{(R^2+R_{s}^2-2 r_{d} R)^{3/2}}\hat{x}[/tex].

    Next, I integrate over the length of a cylinder such that rings at the ends coincide with the surface of the sphere. Noting that r_d is the distance from the center of the sphere to a ring, I can show that the length of a cylinder of a given radius will be 2*r_d. The force is now [tex]\vec{F}=\int_{-r_{d}}^{r_{d}}\frac{2 \pi K n (R-r_{d}\prime)\sqrt{R_{s}^2-r_{d}^2}}{(R^2+R_{s}^2-2 r_{d}\prime R)^{3/2}}\hat{x}dr_{d}\prime[/tex], noting that the r_d in [tex]\sqrt{R_{s}^2-r_{d}^2}[/tex] should not be primed as this represents the radius and is constant for a given cylinder. Also, n is adjusted accordingly. Putting this into MATLAB, I get the solution [tex]\vec{F}=2 \pi K n \hat{x} \frac{\sqrt{R_{s}^2-r_{d}^2}}{R^2}(\frac{r_{d} R-R_{s}^2}{\sqrt{R^2+R_{s}^2-2 r_{d} R}}+\frac{r_{d} R+R_{s}^2}{\sqrt{R^2+R_{s}^2+2 r_{d} R}})[/tex].

    Finally, I integrate over all cylinders in the sphere by integrating the cylinder half-length r_d from 0 to [tex]\sqrt{R_{s}^2-R_{d}^2}[/tex]. This results in the following equation:
    [tex]\vec{F}=2 \pi K n \hat{x} R^{-2}(R \int_{0}^{\sqrt{R_{s}^2-R_{d}^2}} \frac{\sqrt{R_{s}^2-r_{d}^2} r_{d} dr_{d}}{\sqrt{R^2+R_{s}^2-2 r_{d} R}} - R_{s}^2 \int_{0}^{\sqrt{R_{s}^2-R_{d}^2}} \frac{\sqrt{R_{s}^2-r_{d}^2} dr_{d}}{\sqrt{R^2+R_{s}^2-2 r_{d} R}} + R \int_{0}^{\sqrt{R_{s}^2-R_{d}^2}} \frac{\sqrt{R_{s}^2-r_{d}^2} r_{d} dr_{d}}{\sqrt{R^2+R_{s}^2+2 r_{d} R}} + R_{s}^2 \int_{0}^{\sqrt{R_{s}^2-R_{d}^2}} \frac{\sqrt{R_{s}^2-r_{d}^2} dr_{d}}{\sqrt{R^2+R_{s}^2+2 r_{d} R}})[/tex].

    To simplify this equation, I multiply the integrands by [tex]\sqrt{\frac{1/2R}{1/2R}}[/tex], and then let r_d = x, R_s = a, and (R^2+R_s^2)/2R = b. I then find that the integrals are of the following forms:
    [tex]
    1. \int_{A}^{B} \frac{\sqrt{a^2-x^2} x dx}{\sqrt{b-x}} \ \ \
    2. \int_{A}^{B} \frac{\sqrt{a^2-x^2} dx}{\sqrt{b-x}} \ \ \
    3. \int_{A}^{B} \frac{\sqrt{a^2-x^2} x dx}{\sqrt{b+x}} \ \ \
    4. \int_{A}^{B} \frac{\sqrt{a^2-x^2} dx}{\sqrt{b+x}} \ \ \
    [/tex]

    I have not found any solutions to these integrals either by using MATLAB or by looking up integral tables. Nor can I see any other way of constructing the sphere that would properly account for the hole without further complicating the problem. Any advice on how I might move forward would be appreciated.
     
    Last edited: Jan 24, 2009
  2. jcsd
  3. Jan 24, 2009 #2

    gabbagabbahey

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    Gold Member

    Yikes!!!:yuck:

    There is a much easier way to approach this problem. Any force that obeys an inverse square law, also obeys the superposition principle. That means that is you add the forces due to a solid charged sphere of number density [itex]n[/itex] and a cylinder of number density [itex]-n[/itex] that lies inside the sphere exactly like the hole you had before, you get the same thing as the force due to the sphere with a hole in it.
     
  4. Jan 24, 2009 #3

    turin

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    Homework Helper

    I don't think that is true. Can you explain?
     
  5. Jan 24, 2009 #4
    Assuming superposition works*, the cylinder would need spherical end caps to properly coincide with the sphere, otherwise there would be "left over" material and I would find an approximate solution. I suppose I could construct the cylinder in a piecewise manner using integration by disks, remembering to vary the radius of the disks appropriately for the end caps. However, on an earlier attempt at this problem I had tried to construct the complete sphere and hole using integration by disks and found the final integration to be more troublesome than what I found integrating over cylindrical shells, so I'm somewhat reluctant to try integration by disks again. Nevertheless, I'll make the attempt and update accordingly.

    *Superposition should work since two particles at the same location and with equal but opposite charges should cancel each other out. The inverse r-squared law shouldn't even have anything to do with it. The downside is I would have to form the sphere and the cylinder separately, meaning there would be more integrals to solve.

    Update: I've taken the earlier stated equation for a ring and integrated over radius to create a disk, which I then integrated over the diameter of the sphere to create the solid sphere. Specifically, I find the force on a particle due to a solid sphere is
    [tex]
    \vec{F}=\int_{-R_{s}}^{R_{s}}2 \pi K n \hat{x} (1-\frac{R-r_{d}}{\sqrt{R^2+R_{s}^2-2 r_{d} R}}) dr_{d} = \frac{-2 \pi K n \hat{x}}{3 R^2}(2 R^3-3 R_{s} R^2-R_{s}^3-(2 R^3-3 R_{s} R^2+R_{s}^3)csgn(R-R_{s}))
    [/tex]
    where csgn is the complex sign function. Hopefully the integration over the cylinder won't present any extra difficulties.
     
    Last edited: Jan 24, 2009
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