Chapter 17, Problem 77GP

(a)

To determine

The speed of the charge $Q_1$ after very long time if $Q_1$ is released from rest.

Answer to Problem 77GP

The speed of the charge $Q_1$ after it released from the rest is $\text{2}\text{.7}\times {\text{10}}^3\text{m/s}$ .

Explanation of Solution

Given info:

Separation is $r=4\text{cm}$ .

The charge $Q_1$ and $Q_2$ is $5\mu \text{C}$ .

The masses are $m_1=1.5\text{mg}$ and $m_2=2.5\text{mg}$ .

Formula used:

The potential energy due to charges $Q_1$ and $Q_2$ at a distance $r$ is,

  $PE=\frac{k{Q}_1{Q}_2}{r}$

The kinetic energies are,

  $\begin{array}{c}K{E}_1=\frac{1}{2}{m}_1{v}_1^2\\ K{E}_2=\frac{1}{2}{m}_2{v}_2^2\end{array}$

If the charge $Q_1$ is released from the rest then its kinetic energy will be equal to its potential energy.

  $\begin{array}{c}K{E}_1=PE\\ \frac{1}{2}{m}_1{v}_1^2=\frac{k{Q}_1{Q}_2}{r}\\ v_1=\sqrt{\frac{2k{Q}_1{Q}_2}{m_{1}r}}\end{array}$

Here, k is the Coulomb’s constant.

Calculation:

Substituting the given values,

  $\begin{array}{c}{v}_1=\sqrt{\frac{2\left(9\times {10}^9\right)\left(5\times {10}^{-6}\right)\left(\left(5\times {10}^{-6}\right)\right)}{\left(1.5\times {10}^{-6}\right)\left(4\times {10}^{-2}\right)}}\text{m/s}\\ =\text{2}\text{.7}\times {\text{10}}^3\text{m/s}\end{array}$

Conclusion:

Thus, the speed of the charge $Q_1$ after it released from the rest is $\text{2}\text{.7}\times {\text{10}}^3\text{m/s}$ .

(b)

To determine

The speed of the charge $Q_1$ after very long time if both charges are released from rest.

Answer to Problem 77GP

The speed of the charge $Q_1$ after very long time if both charges are released from rest is $\text{2}\text{.2}\times {\text{10}}^3\text{m/s}$ .

Explanation of Solution

Given info:

Separation is $r=4\text{cm}$ .

The charge $Q_1$ and $Q_2$ is $5\mu \text{C}$ .

The masses are $m_1=1.5\text{mg}$ and $m_2=2.5\text{mg}$ .

Formula used:

Initially, both point charges are released from the rest. Hence, the initial momentum of the point charges is zero.

  $p_{\text{initial}}=0$

The final momentums of the two point charges are,

  $p_{\text{final}}=m_1{v}_1+m_2{v}_2$

In this case, both momentum and the energy of the system will be conserved. Since, the initial momentum is zero, and the magnitudes of the moments of the two charges will be equal.

  $\begin{array}{c}{p}_{\text{inital}}=p_{\text{final}}\\ 0=m_1{v}_1+m_2{v}_2\\ v_2=-v_1\frac{m_1}{m_2}\end{array}$

From the conservation of Energy, the potential energy of the two point charges will be equal to the kinetic energy of the two point charges.

  $\begin{array}{c}K{E}_{\text{final}}=PE\\ PE=K{E}_1+K{E}_2\\ \frac{k{Q}_1{Q}_2}{r}=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\\ =\frac{1}{2}\left(m_1{v}_1^2+m_2{v}_2^2\right)\end{array}$

Replace $v_2=-v_1\frac{m_1}{m_2}$ .

  $\begin{array}{c}\frac{k{Q}_1{Q}_2}{r}=\frac{1}{2}\left(m_1{v}_1^2+m_2{\left(-v_1\frac{m_1}{m_2}\right)}^2\right)\\ =\frac{1}{2}{m}_1{v}_1^2\left(\frac{m_2+m_1}{m_2}\right)\\ v_1=\sqrt{\frac{2k{Q}_1{Q}_2{m}_2}{r{m}_1\left(m_1+m_2\right)}}\end{array}$

Calculation:

Substituting the given values,

  $\begin{array}{c}{v}_1=\sqrt{\frac{2\left(9\times {10}^9\right)\left(5\times {10}^{-6}\right)\left(5\times {10}^{-6}\right)\left(2.5\times {10}^{-6}\right)}{\left(1.5\times {10}^{-6}\right)\left(4\times {10}^{-2}\right)\left(1.5\times {10}^{-6}+2.5\times {10}^{-6}\right)}}\text{m/s}\\ =\text{2}\text{.2}\times {\text{10}}^3\text{m/s}\end{array}$

Conclusion:

Thus, the speed of the charge $Q_1$ after very long time if both charges are released from rest is $\text{2}\text{.2}\times {\text{10}}^3\text{m/s}$ .