Chapter 26, Problem 60PQ

(a)

To determine

The speed of orbiting particle.

Answer to Problem 60PQ

The speed of orbiting particle is $17.0\text{m/s}$.

Explanation of Solution

Write an expression for the culoumb force.

  $F=\frac{k\left|qQ\right|}{r^2}$                                                                                                           (I)

Here, $k$ is the culoumb constant, $Q$ is charge of the fixed particle, $q$ is charge of moving particle, $r$ is the distance between the sharges and $F$ is the electric potential.

Write an expression for the centripetal force.

  $F_C=\frac{m{v}^2}{r}$                                                                                                           (II)

Here, $m$ is mass of moving particle, $v$ is the speed of particle and $F_C$ is the centripetal force.

The centripetal force and culoumb force are balances to keep the particle in motion.

Equate equation (I) and (II).

  $\frac{k\left|qQ\right|}{r^2}=\frac{m{v}^2}{r}$                                                                                                    (III)

Rearrange equation (III) to find $v$.

  $v=\sqrt{\frac{k\left|qQ\right|}{mr}}$                                                                                                      (IV)

Conclusion:

Substitute $8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $2.7\text{nC}$ for $q$, $-8.6\text{μC}$ for $Q$, $0.57\text{g}$ for $m$ and $1.3\text{mm}$ for $r$ in equation (IV) to find $V$.

    $\begin{array}{c}v=\sqrt{\frac{\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left|\left(\left(2.7\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)\left(\left(-8.6\text{μC}\right)\left(\frac{1\text{C}}{{10}^6\text{μC}}\right)\right)\right|}{\left(\left(0.57\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)\left(\left(1.3\text{mm}\right)\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}}\\ =\sqrt{\frac{\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left|\left(2.7\times {10}^{-9}\text{C}\right)\left(-8.6\times {10}^{-6}\text{C}\right)\right|}{\left(0.57\times {10}^{-3}\text{kg}\right)\left(1.3\times {10}^{-3}\text{m}\right)}}\\ =17.0\text{m/s}\end{array}$

Thus, the speed of orbiting particle is $17.0\text{m/s}$.

(b)

To determine

Electric potential energy of the system.

Answer to Problem 60PQ

Electric potential energy of the system is $-0.16\text{J}$.

Explanation of Solution

Write an expression for the electric potential energy of the system.

  $U_E=\frac{kQq}{r}$                                                                                                          (V)

Here, $U_E$ is the electric potential energy of the system.

Conclusion:

Substitute $8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $2.7\text{nC}$ for $q$, $-8.6\text{μC}$ for $Q$ and $1.3\text{mm}$ for $r$ in equation (V) to find $U_E$.

    $\begin{array}{c}{U}_E=\frac{\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(\left(-8.6\text{μC}\right)\left(\frac{1\text{C}}{{10}^6\text{μC}}\right)\right)\left(\left(2.7\text{nC}\right)\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{\left(\left(1.3\text{mm}\right)\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}\\ =\frac{\left(8.99\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(-8.6\times {10}^{-6}\text{C}\right)\left(2.7\times {10}^{-9}\text{C}\right)}{\left(1.3\times {10}^{-3}\text{m}\right)}\\ =-0.16\text{J}\end{array}$

Thus, the electric potential energy of the system is $-0.16\text{J}$.

(c)

To determine

Total energy of the system.

Answer to Problem 60PQ

The total energy of the system is $-0.078\text{J}$.

Explanation of Solution

Write an expression for the kinetic energy.

  $K=\frac{1}{2}m{v}^2$                                                                                                         (VI)

Here, $K$ is the kinetic energy.

Write an expression for the total energy.

  $E=U_E+K$                                                                                                     (VII)

Here, $E$ is the total energy.

Conclusion:

Substitute $0.57\text{g}$ for $m$ and $17.0\text{m/s}$ for $v$ in equation (VI) to find $K$.

    $\begin{array}{c}K=\frac{1}{2}\left(\left(0.57\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{mm}}\right)\right){\left(17.0\text{m/s}\right)}^2\\ =\frac{1}{2}\left(0.57\times {10}^{-3}\text{kg}\right){\left(17.0\text{m/s}\right)}^2\\ =0.082\text{J}\end{array}$

Substitute $0.082\text{J}$ for $K$ and $-0.16\text{J}$ for $U_E$ in equation (VII) to find $E$.

    $\begin{array}{c}E=0.082\text{J}+\left(-0.16\text{J}\right)\\ =0.082\text{J}-0.16\text{J}\\ =-0.078\text{J}\end{array}$

Thus, the total energy of the system is $-0.078\text{J}$.