Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 20 - Electric Potential and Electric Potential Energy - Problems and Conceptual Exercises - Page 719: 39

Answer

(a) $76.7KV $ (b) $14.1m/s $

Work Step by Step

(a) We can find the distances $ r_1, r_3$ as $ r_1,r_3=\frac{1.25m}{2}=0.625m $ and $ r_2=\sqrt{(1.25m)^2-(0.625m)^2}=1.083m $ Now $ V=K(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3})$ We plug in the known values to obtain: $ V=(8.99\times 10^9N.m^2/C^2)[\frac{2.75}{0.625m}+\frac{7.45}{1.083m}+\frac{(-1.72)}{0.625m}]\times 10^{-6}C=76.7KV $ (b) According to the law of conservation of energy $ K_i+U_i=K_i+U_f $ $\implies 0+q_4V=\frac{1}{2}mv_f^2+0$ This simplifies to: $ v_f=\sqrt{\frac{2q_4V}{m}}$ We plug in the known values to obtain: $ v_f=\sqrt{\frac{2(6.11\times 10^{-6}C)(7.67\times 10^3V)}{4.71\times 10^{-3}Kg}}$ $ v_f=14.1m/s $
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