Question

Q) A small, stationary sphere carries a net charge Q. You perform the following experiment to measure Q: From a large distance you fire a small particle with mass m = 4.00 x 10-4 kg and charge q = 5.00 x 10-8 C directly at the center of the sphere. The apparatus you are using measures the particle’s speed v as a function of the distance x from the sphere. The sphere’s mass is much greater than the mass of the projectile particle, so you assume that the sphere remains at rest. All of the measured values of x are much larger than the radius of either object, so you treat both objects as point particles. You plot your data on a graph of v2 versus (1/x) (Fig. P23.80). The straight line v2 = 400 m2/s2 – [15.75 m3/s2)/x] gives a good fit to the data points. (a) Explain why the graph is a straight line. (b) What is the initial speed v0 of the particle when it is very far from the sphere? (c) What is Q? (d) How close does the particle get to the sphere? Assume that this distance is much larger than the radii of the particle and sphere, so continue to treat them as point particles and to assume that the sphere remains at rest.

Short answer

(a) The square of the final speed ${\mathrm{v}}_{\mathrm{f}}^2$is directly proportional to $1/\mathrm{r}$. So, the graph is a straight line with slope $-\frac{1}{2{\mathrm{\pi \epsilon }}_{\circ }}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{m}}$.

(b) The initial speed vi of the particle when it is very far from the sphere is 20 m/s.

(c) The value of Q is$7.008\times {10}^{-6}\mathrm{C}$

(d) The particle is closest to the sphere at 0.039 m.

Step-by-step solution

Step 1:

According to conservation law of energy:

$$\begin{gathered} \mathbf{\Delta KE}\mathbf{-}\mathbf{\Delta U}\mathbf{=}\mathbf{0} \\ {\mathbf{K}}_{\mathbf{f}}\mathbf{-}{\mathbf{K}}_{\mathbf{i}}\mathbf{=}{\mathbf{U}}_{\mathbf{i}}\mathbf{-}{\mathbf{U}}_{\mathbf{f}} \end{gathered}$$

Potential energy is

$\begin{array}{l}\mathrm{U}=\mathrm{qV}\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_{°}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{r}}\end{array}$

Substitute the values to get:

role="math" localid="1664871622974" $$\begin{gathered} \frac{1}{2}m{v_i}^2-\frac{1}{2}m{v^2}_f=\frac{1}{4{\mathrm{\pi \epsilon }}_{°}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{r}}-0 \\ {v_i}^2-{v^2}_f=\frac{1}{4{\mathrm{\pi \epsilon }}_{°}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{rm}} \\ ={v_i}^2-{v^2}_f=\frac{1}{2{\mathrm{\pi \epsilon }}_{°}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{rm} \end{gathered}$$

The square of the final speed ${v^2}_f$is directly proportional to $\frac{1}{r}$ . So, the graph is a straight line with slope $\frac{1}{2{\mathrm{\pi \epsilon }}_{°}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{r}\mathrm{m}}$.

Step 2:

Since the particle is very far from the sphere, then the distance between them is∞. Substitute in the previous calculation:

$$\begin{gathered} {{\mathrm{v}}_{\mathrm{i}}}^2={{\mathrm{v}}^2}_{\mathrm{f}}-\frac{1}{2{\mathrm{\pi \epsilon }}_{°}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{r}\mathrm{m}} \\ ={{\mathrm{v}}^2}_{\mathrm{f}}-\frac{1}{2{\mathrm{\pi \epsilon }}_{°}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{r}\mathrm{m}} \\ ={{\mathrm{v}}^2}_{\mathrm{f}} \\ {\mathrm{v}}_{\mathrm{f}}=\sqrt{{{\mathrm{v}}^2}_{\mathrm{f}}} \\ =\sqrt{400{\mathrm{m}}^2/\mathrm{s}} \\ =20\mathrm{m}/\mathrm{s} \end{gathered}$$

The initial speed vi of the particle when it is very far from the sphere is 20 m/s.

Step 3:

Substitute 15.75 m3/s2for the slope$-\frac{1}{2{\mathrm{\pi \epsilon }}_{°}}\frac{\mathrm{Qq}}{\mathrm{m}}$

$$\begin{gathered} -\frac{1}{2{\mathrm{\pi \epsilon }}_{°}}\frac{\mathrm{Qq}}{\mathrm{m}}=-15.75m^3/s^2 \\ Q=\frac{-m(-15.75m^3/s^2)}{q} \\ =\frac{\left(4.00\times {10}^4\right)\left(15.75m^3/s^2\right)}{2\left(9.00\times {10}^{9}N{m}^2/C^2\right)\left(5.00\times {10}^{-9}C\right)} \\ =7.008\times {10}^{-6}C \end{gathered}$$

The value of Q is$7.008\times {10}^{-6}C$

Step 4:

The particle is closest to sphere when vf=0

Substitute $-\frac{1}{2{\mathrm{\pi \epsilon }}_{°}}\frac{\mathrm{Qq}}{\mathrm{m}}=-15.75{\mathrm{m}}^3/{\mathrm{s}}^2$to get x:

$v_f=20m/s$

${{\mathrm{v}}_f}^2={{\mathrm{v}}^2}_i-\frac{1}{2{\mathrm{\pi \epsilon }}_{°}}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{x\mathrm{m}}$

role="math" localid="1664873076802" $$\begin{gathered} 0=400{\mathrm{m}}^2/{\mathrm{s}}^2-\frac{15.75{\mathrm{m}}^3/{\mathrm{s}}^2}{\mathrm{x}} \\ x=\frac{15.75{\mathrm{m}}^3/{\mathrm{s}}^2}{400{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =3.94\times {10}^{-2}\mathrm{m} \end{gathered}$$

The particle is closest to the sphere at 0.039 m.