Question

Assume that a ball of charged particles has a uniformly distributednegative charge density except for a narrow radial tunnel throughits center, from the surface on one side to the surface on the opposite side. Also assume that we can position a proton any where along the tunnel or outside the ball. Let \(\displaystyle{F}_{{R}}\) be the magnitude of the electrostatic force on the proton when it islocated at the ball's surface, at radius R. As a multiple ofR, how far from the surface is there a point where the forcemagnitude is 0.44FR if we move the proton(a) away from the ball and (b) into the tunnel?

Answer 1

By Gauss' Law a spherically symmetric charge distribution willcontribute an electric field only outside an imaginary spherical surface drawn at a radiusr within the charge distribution. Any charge outside this imaginary surface does not contributeto the electric field within the surface. Also, the electric field outside such an imaginarysurface is the same as though all of the charge within the surface was concentrated at the center ofthe sphere.
So \(\displaystyle{E}={\frac{{{k}{Q}}}{{{r}^{{2}}}}}\) at a distance r from the center of the spherewhere Q the total charge contained within a spherical surface of radius r is \(\displaystyle{Q}={\frac{{{4}}}{{{3}}}}\cdot\pi\cdot{r}^{{3}}\)
Also, you can write \(\displaystyle{q}={Q}\cdot{\frac{{{4}}}{{{3}}}}\cdot\pi\cdot{\frac{{{r}^{{3}}}}{{{\frac{{{4}}}{{{3}}}}\cdot\pi\cdot{R}^{{3}}}}}={Q}{\frac{{{r}^{{3}}}}{{{R}^{{3}}}}}\) where Q is total charge on the sphere and q is the charge within an imaginary surfaceof radius r. Also, you can write the electric field as \(\displaystyle{E}={\frac{{{Q}{e}}}{{{4}\pi\epsilon_{{0}}{R}^{{3}}}}}\) where Q is total charge, r the distancefrom center of the sphere, and R is the radius of the spherical charge distribution.
Note that r > R you simply have \(\displaystyle{E}={\frac{{{Q}}}{{{4}\pi\epsilon_{{0}}{r}^{{2}}}}}\) where r is distance to center of sphere. Also, note that since the force within the sphere isproportional r, a charged particle released at any point within the tunnel would simple oscillate with simpleharmonic motion with the amplitude that it had when released.

Answer 2

(a) The flux is still \(-750N m^2/C^2\) , since it depends only on the amount of charge enclosed. 

(b) We use  \(\phi=\frac{q}{\epsilon_0}\)​ to obtain the charge q :

\(q=\epsilon_0 \phi =(8.85 \times 10^{-12} C^2 / N \ m^2)(-750 \ N. m^2/C^2)=\)

\(-6.64 \times 10^{-9} C\)