Chapter 24, Problem 36P

To determine

The speed of the proton.

Answer to Problem 36P

The speed of the proton is $5.94\times {10}^5\text{m}/\text{s}$.

Explanation of Solution

Write the expression for the total enclosed charge.

    $Q_{\text{enc}}=\left(\rho \frac{4}{3}\pi \left(r_1^3-r_2^3\right)\right)+q$                                                                                       (I)

Here, $Q_{\text{enc}}$ is the total enclosed charge, $\rho$ is the volume charge density, $r_1$ and $r_2$ is the outer and inner radius respectively and $q$ is charge of the particle.

Write the expression for electric field force on the proton.

    $F=eE$                                                                                                                     (II)

Here, $F$ is the electric field force on the proton, $e$ is the charge of the proton and $E$ is the electric field.

Substitute $\frac{Q_{\text{enc}}}{4\pi {\epsilon }_0{r}_1^2}$ for $E$ in equation (II) to calculate $F$.

    $F=e\left(\frac{Q_{\text{enc}}}{4\pi {\epsilon }_0{r}_1^2}\right)$                                                                                                      (III)

Write the expression for centripetal force acting on the proton.

    $F=\frac{m_{\text{p}}\times v^2}{r_1}$                                                                                                             (IV)

Here, $m_{\text{p}}$ is the mass of the proton, $v$ is the velocity of the proton, $r_1$ is the orbit radius, $F$ is the centripetal force acting on the proton.

From Equation (III) and Equation (IV),

    $e\left(\frac{Q_{\text{enc}}}{4\pi {\epsilon }_0{r}_1^2}\right)=\frac{m_{\text{p}}{v}^2}{r_1}$

Rewrite the above equation for $v$.

    $\begin{array}{c}e\left(\frac{Q_{\text{enc}}}{4\pi {\epsilon }_0{r}_1^2}\right)=\frac{m_{\text{p}}{v}^2}{r_1}\\ v^2=e\left(\frac{r_1}{m_{\text{p}}}\right)\left(\frac{Q_{\text{enc}}}{4\pi {\epsilon }_0{r}_1^2}\right)\\ v=\sqrt{e\left(\frac{r_1}{m_{\text{p}}}\right)\left(\frac{k{Q}_{\text{enc}}}{r_1^2}\right)}\end{array}$                                                                                (V)

Conclusion:

Substitute $-1.33\mu \text{C}/{\text{m}}^{\text{3}}$ for $\rho$, $20.0\text{cm}$ for $r_2$, $25.0\text{cm}$ for $r_1$ and $-60.0\text{nC}$ for $q$ in equation (I) to calculate $Q_{\text{enc}}$.

    $\begin{array}{c}{Q}_{\text{enc}}=\left(-1.33\mu \text{C}/{\text{m}}^{\text{3}}\right)\left(\frac{4}{3}\right)\pi \left[{\left(25.0\text{cm}\right)}^3-{\left(20.0\text{cm}\right)}^3\right]+\left(-60.0\text{nC}\right)\\ \text{=}\left[\begin{array}{l}\left(\left(-1.33\mu \text{C}/{\text{m}}^{\text{3}}\right)\left(\frac{1\text{C}}{{10}^6\mu \text{C}}\right)\times \left(\frac{4}{3}\right)\pi \times \left(15625{\text{cm}}^3-8000{\text{cm}}^3\right){\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)}^3\right)\\ +\left(-60\text{nC}\right)\left(\frac{1\times {10}^{-9}\text{C}}{\text{1}\text{nC}}\right)\end{array}\right]\\ =\left(\left(-1.33\times {10}^{-6}\text{C}/{\text{m}}^{\text{3}}\right)\left(\frac{4}{3}\right)\pi \left(7.625\times {10}^{-3}{\text{m}}^3\right)\right)+\left(-60\times {10}^{-9}\text{C}\right)\\ =\left(-0.0425\times {10}^{-6}\text{C}\right)+\left(-60\times {10}^{-9}\text{C}\right)\end{array}$

Further solve the above equation.

    $Q_{\text{enc}}=-1.025\times {10}^{-7}\text{C}$

Substitute $1.6\times {10}^{-19}\text{C}$ for $e$, $1.025\times {10}^{-7}\text{C}$ for $Q_{\text{enc}}$, $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $1.67\times {10}^{-27}\text{Kg}$ for $m_{\text{p}}$, $0.25\text{m}$ for $r_1$ in equation (V) to calculate $v$.

    $\begin{array}{c}v=\sqrt{\left(1.6\times {10}^{-19}\text{C}\right)\frac{\left(1.025\times {10}^{-7}\text{C}\right)\left(0.25\text{m}\right)\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)}{\left(1.67\times {10}^{-27}\text{kg}\right){\left(0.25\text{m}\right)}^2}}\\ =\sqrt{\left(1.6\times {10}^{-19}\text{C}\right)\frac{230.625\text{N}\cdot {\text{m}}^3/\text{C}}{1.044\times {10}^{-28}\text{kg}\cdot {\text{m}}^2}}\\ =\sqrt{35.35\times {10}^{10}{\text{m}}^2/{\text{s}}^2}\\ =5.94\times {10}^5\text{m}/\text{s}\end{array}$

Therefore, the velocity of the proton is $5.94\times {10}^5\text{m}/\text{s}$.