Question

Assume that a ball of charged particles has a uniformly distributed negative charge density except for a narrow radial tunnel through its center, from the surface on one side to the surface on the opposite side. Also assume that we can position a proton anywhere along the tunnel or outside the ball. Let ${\mathit{F}}_{\mathbf{r}}$ be the magnitude of the electrostatic force on the proton when it is located at the ball’s surface, at radius R. As a multiple of R, how far from the surface is there a point where the force magnitude is if we move the proton (a) away from the ball and (b) into the tunnel?

Short answer

a) As a multiple of R , the point is at a distance of 0.41 R where the magnitude of the force is $0.50{\mathrm{F}}_{\mathrm{R}}$if we move the proton away from the ball.

b) As a multiple of , the point is at a distance of 0.50 R where the magnitude of the force is $0.50{\mathrm{F}}_{\mathrm{R}}$if we move the proton into the tunnel.

Step-by-step solution

The given data

a) The magnitude of the electrostatic force on the proton when it is located at the ball’s surface, at radius R , is ${\mathrm{F}}_{\mathrm{R}}$.

b) The magnitude of the force is 0.50${\mathrm{F}}_{\mathrm{R}}$ .

Understanding the concept of the electric field and the electrostatic force

Using the concept of the electric field in the formula of the electrostatic force, we can get the required magnitudes of the force. Thus, using this value we can calculate the values of the distance of the point.

Formulae:

The electric field at a point due to a particle charge, $\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}$ (1)

The electric field of a point inside the surface, $\mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{qr}}{{\mathrm{R}}^3}$ (2)

The electrostatic force on a charged particle, $\mathrm{F}=\mathrm{qE}$ (3)

a) Calculation of the distance of the point from the surface when the proton is away from the ball

The field at the proton’s location (but not caused by the proton) has magnitude E . The proton’s charge is e . The ball’s charge has magnitude q . Thus, as long as the proton is at $\mathrm{r}\ge \mathrm{R}$then the force on the proton (caused by the ball) has magnitude by substituting the given values and equation (1) in equation (3) as follows:

$$\begin{gathered} \mathrm{F}=\mathrm{e}\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}\right) \\ =\frac{\mathrm{eq}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} \end{gathered}$$

where, r is measured from the center of the ball (to the proton). We note that if r = R , then the above expression becomes

${\mathrm{F}}_{\mathrm{R}}=\frac{\mathrm{eq}}{4{\mathrm{\pi \epsilon }}_0{R}^2}$

If we require, the magnitude of the force to be$\mathrm{F}=\frac{1}{2}{\mathrm{F}}_{\mathrm{R}}$, then using equation (1) and the above value we get the distance of the point from the center of the sphere as:

.role="math" localid="1657358693702" $$\begin{gathered} \frac{\mathrm{eq}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}=\frac{1}{2}\left(\frac{\mathrm{eq}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2}\right) \\ \mathrm{r}=\sqrt{2}\mathrm{R} \end{gathered}$$

Now, the distance from the surface is given as:

$\sqrt{2}\mathrm{R}-\mathrm{R}=0.41\mathrm{R}$

Hence, the value of the distance of the point is 0.41 R .

b) Calculation of the distance of the point from the surface when the proton is into the tunnel

Now, we require the force of the magnitude as: ${\mathrm{F}}_{\mathrm{inside}}=\frac{1}{2}{\mathrm{F}}_{\mathrm{R}}$

where ${\mathrm{F}}_{\mathrm{inside}}={\mathrm{eE}}_{\mathrm{inside}}$and ${\mathrm{E}}_{\mathrm{inside}}$is given.

Thus, the distance value from the surface when the proton goes into the tunnel is given using equation (2) and equation (3) as:

$$\begin{gathered} \mathrm{e}\left(\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^3}\right)\mathrm{r}=\frac{1}{2}\frac{\mathrm{eq}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2} \\ \mathrm{r}=0.50\mathrm{R} \end{gathered}$$

Hence, the value of the distance is 0.50 R .