Force on Particle due to Sphere with Cylindrical Hole

[Leyic]

Note: This is my first time using LaTeX. Any formatting advice would be appreciated.

Homework Statement

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 \vec{F}=\frac{K}{r^2}\hat{r}, 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.

Homework Equations

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 \vec{F} = \frac{2 K r_{x}}{r^3}\hat{x}, with r=\sqrt{r_{x}^2+r_{y}^2} 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, \vec{F}=\frac{2 K r_{x}}{r^3}\hat{x} \pi r_{y} n, where n is a number density term. I then note that the distance from the center of the sphere to the center of the ring is r_{d}=\sqrt{R_{s}^2+r_{y}^2} and that r_x = R - r_d. I can then rewrite the force caused by the ring as \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}.

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 \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, noting that the r_d in \sqrt{R_{s}^2-r_{d}^2} 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 \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}}).

Finally, I integrate over all cylinders in the sphere by integrating the cylinder half-length r_d from 0 to \sqrt{R_{s}^2-R_{d}^2}. This results in the following equation:
\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}}).

To simplify this equation, I multiply the integrands by \sqrt{\frac{1/2R}{1/2R}}, 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:
<br /> 1. \int_{A}^{B} \frac{\sqrt{a^2-x^2} x dx}{\sqrt{b-x}} \ \ \<br /> 2. \int_{A}^{B} \frac{\sqrt{a^2-x^2} dx}{\sqrt{b-x}} \ \ \<br /> 3. \int_{A}^{B} \frac{\sqrt{a^2-x^2} x dx}{\sqrt{b+x}} \ \ \<br /> 4. \int_{A}^{B} \frac{\sqrt{a^2-x^2} dx}{\sqrt{b+x}} \ \ \<br />

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.

 

[gabbagabbahey]

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 \vec{F} = \frac{2 K r_{x}}{r^3}\hat{x}, with r=\sqrt{r_{x}^2+r_{y}^2} 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, \vec{F}=\frac{2 K r_{x}}{r^3}\hat{x} \pi r_{y} n, where n is a number density term.

Yikes!

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 n and a cylinder of number density -n 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.

 

[turin]

Any force that obeys an inverse square law, also obeys the superposition principle.

I don't think that is true. Can you explain?

 

[Leyic]

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 n and a cylinder of number density -n 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.

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
<br /> \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}))<br />
where csgn is the complex sign function. Hopefully the integration over the cylinder won't present any extra difficulties.