here,
charge , Q1 = 3.4 * 10^-6 C is at x1 = 62 cm = 0.62 m
a)
when Q = 55 * 10^-6 C
r = 0.16 m
let the kinetic energy be KE
KE = K * Q * Q1 * ( 1 /x1 - 1/(x1 + r))
KE = 9 * 10^9 * 3.4 * 10^-6 * 55 * 10^-6 * ( 1/0.62 - 1/0.78)
KE = 0.56 J
b)
when Q = - 55 * 10^-6 C
r = 0.16 m
let the kinetic energy be KE
KE = K * Q * Q1 * ( 1 /(x1 - r ) - 1/(x1))
KE = 9 * 10^9 * 3.4 * 10^-6 * 55 * 10^-6 * ( 1/(0.62 - 0.16) - 1/0.62)
KE = 0.94 J
Chapter 24, Probie 3/15/2018 Print by: MELISSA SIMION Physics 202-001 Spring 2018/ DiFabio Phys 202 Chap...
Halliday, Fundamentals of Physics, 10e Calculus-based Physics I & II (PH 201-202) tice Assignment Gradebook ORION Downloadable eTextbook Assignment FULL SCREEN PRINTER VERSION BACK Chapter 21, Problem 022 The figure shows an arrangement of four charged particles, with angle θ-40.0 ° and distance d-2.50 cm. Particle 2 has charge q2 960x 10-19 С; particles 3 and 4 have charges q3-q,--3.20 x 10-19 C. (a) What is the distance D between the origin and particle 2 if the net electrostatic force...
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