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# Assume that a ball of charged particles has a uniformly distributednegative charge density except for a narrow radial tunnel throughits center, from t

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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?

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.