Cutnell, Physics, 10e Help I S 1-EI PRINTER VERSION BACK NEXT Chapter 19, Proble

Question

Cutnell, Physics, 10e Help I S 1-EI PRINTER VERSION BACK NEXT Chapter 19, Problem 10 A moving particle encounters an external electric field that decreases its kinetic energy from 9850 eV to 7120 eV as the particle moves from position A to position B. The electric potential at A is-69.0 V, and that at B is +19.0 V. Determine the charge of the particle. Include the algebraic sign(+ or -) with your answer Higher potential Lower potential Units Number the tolerance is +/-296 Click if you would like to Show Work for this question: Open Show Work

Explanation / Answer

change in kinetic energy = 7120-9850 = -2730 eV = -2730*1.6*10^-19 = -43.68*10^-17 J

from conservation energy

potential energy = change in kinetic energy

q(VB-VA) = -43.68*10^-17

q(19-(-69)) = -43.68*10^-17

q = -4.96*10^-18 C

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.