Solved: Two point charges are moving to the right along | StudySoup

Textbook Solutions for Sears and Zemansky's University Physics with Modern Physics

Chapter 23 Problem 23.92

Question

Two point charges are moving to the right along the x-axis. Point charge 1 has charge \(q_1=2.00\ \mu\mathrm{C}\), mass \(m_1 = 6.00 \times 10^{-5}\mathrm{\ kg}\), and speed \(v_{1}\). Point charge 2 is to the right of \(q_{1}\) and has charge \(q_2 = -5.00\ \mu\mathrm{C}\), mass \(m_2 = 3.00 \times 10^{-5}\mathrm{\ kg}\), and speed \(v_{2}\). At a particular instant, the charges are separated by a distance of and have speeds \(v_1 = 400\ \mathrm{m}/\mathrm{s}\quad\text{ and } v_2 = 1300 \mathrm{\ m}/\mathrm{s}\). The only forces on the particles are the forces they exert on each other. (a) Determine the speed \(v_{\mathrm{cm}}\) of the center of mass of the system. (b) The relative energy \(E_{\mathrm{rel}}\) of the system is defined as the total energy minus the kinetic energy contributed by the motion of the center of mass:

\(E_{\mathrm{rel}}=E-\frac{1}{2}\left(m_{1}+m_{2}\right) v_{\mathrm{cm}}^{2}\)

where \(E=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}+q_{1} q_{2} / 4 \pi \epsilon_{0} r\) is the total energy of the system and r is the distance between the charges. Show that \(E_{\text {rel }}=\frac{1}{2} \mu v^{2}+q_{1} q_{2} / 4 \pi \epsilon_{0} r, \text { where } \mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)\) is called the reduced mass of the system and \(v=v_{2}-v_{1}\) is the relative speed of the moving particles. (c) For the numerical values given above, calculate the numerical value of \(E_{\mathrm{rel}}\). (d) Based on the result of part (c), for the conditions given above, will the particles escape from one another? Explain. (e) If the particles do escape, what will be their final relative speed when \(r \rightarrow \infty\)? If the particles do not escape, what will be their distance of maximum separation? That is, what will be the value of r when v = 0? Repeat parts (c)–(e) for \(v_1 = 400 \mathrm{\ m}/\mathrm{s} \text { and } v_2 = 1800 \mathrm{\ m}/\mathrm{s}\) when the separation is 9.00 mm.

Text Transcription:

q_1=2.00 mu C

m_1 = 6.00 times 10^-5 kg

v_1

q_1

q_2 = -5.00 mu C

m_2 = 3.00 times 10^-5 kg

v_2

v_1 = 400 m/s and v_2 = 1300 m/s

v_cm

E_rel

E_rel = E-1/2(m_1+m_2)v_cm^2

E = 1/2 m_1v_1^2 + 1/2 m_2v_2^2 + q_1q_2/4 pi epsilon_0 r

E_rel = 1/2 mu v^2 + q_1q_2/4 pi epsilon_0 r, where mu = m_1m_2/(m_1 + m_2)

v = v_2 - v_1

r rightarrow infty

v_1 = 400 m/s and v_2 = 1800 m/s

Solution

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The first step in solving 23 problem number 92 trying to solve the problem we have to refer to the textbook question: Two point charges are moving to the right along the x-axis. Point charge 1 has charge \(q_1=2.00\ \mu\mathrm{C}\), mass \(m_1 = 6.00 \times 10^{-5}\mathrm{\ kg}\), and speed \(v_{1}\). Point charge 2 is to the right of \(q_{1}\) and has charge \(q_2 = -5.00\ \mu\mathrm{C}\), mass \(m_2 = 3.00 \times 10^{-5}\mathrm{\ kg}\), and speed \(v_{2}\). At a particular instant, the charges are separated by a distance of and have speeds \(v_1 = 400\ \mathrm{m}/\mathrm{s}\quad\text{ and } v_2 = 1300 \mathrm{\ m}/\mathrm{s}\). The only forces on the particles are the forces they exert on each other. (a) Determine the speed \(v_{\mathrm{cm}}\) of the center of mass of the system. (b) The relative energy \(E_{\mathrm{rel}}\) of the system is defined as the total energy minus the kinetic energy contributed by the motion of the center of mass:\(E_{\mathrm{rel}}=E-\frac{1}{2}\left(m_{1}+m_{2}\right) v_{\mathrm{cm}}^{2}\)where \(E=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}+q_{1} q_{2} / 4 \pi \epsilon_{0} r\) is the total energy of the system and r is the distance between the charges. Show that \(E_{\text {rel }}=\frac{1}{2} \mu v^{2}+q_{1} q_{2} / 4 \pi \epsilon_{0} r, \text { where } \mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)\) is called the reduced mass of the system and \(v=v_{2}-v_{1}\) is the relative speed of the moving particles. (c) For the numerical values given above, calculate the numerical value of \(E_{\mathrm{rel}}\). (d) Based on the result of part (c), for the conditions given above, will the particles escape from one another? Explain. (e) If the particles do escape, what will be their final relative speed when \(r \rightarrow \infty\)? If the particles do not escape, what will be their distance of maximum separation? That is, what will be the value of r when v = 0? Repeat parts (c)–(e) for \(v_1 = 400 \mathrm{\ m}/\mathrm{s} \text { and } v_2 = 1800 \mathrm{\ m}/\mathrm{s}\) when the separation is 9.00 mm.Text Transcription:q_1=2.00 mu Cm_1 = 6.00 times 10^-5 kgv_1q_1q_2 = -5.00 mu Cm_2 = 3.00 times 10^-5 kgv_2v_1 = 400 m/s and v_2 = 1300 m/sv_cmE_relE_rel = E-1/2(m_1+m_2)v_cm^2E = 1/2 m_1v_1^2 + 1/2 m_2v_2^2 + q_1q_2/4 pi epsilon_0 rE_rel = 1/2 mu v^2 + q_1q_2/4 pi epsilon_0 r, where mu = m_1m_2/(m_1 + m_2)v = v_2 - v_1r rightarrow inftyv_1 = 400 m/s and v_2 = 1800 m/s
From the textbook chapter Electric Potential you will find a few key concepts needed to solve this.

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Title Sears and Zemansky's University Physics with Modern Physics 13 
Author Hugh D. Young; Roger A. Freedman; A. Lewis Ford
ISBN 9780321696861

Solved: Two point charges are moving to the right along

Chapter 23 textbook questions

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