In the figure below, four particles are fixed along an x axis, separated by distances d...
In the figure below, four particles are fixed along an x axis, separated by distances d = 2.10 cm. The charges are q1 = +4e, q2 = -e, q3 = +e, and q4 = +5e, with e = 1.60
Homework Answers
net force of q1 = F12 + F13 +F14
F12 = kq1q2/d^2 = 9e9 * 4*1.6 e-19 *1.6 e-19 /(0.021*0.021)
F12 = -2.1 e-24 N
F13 = kq1q3/(2d)^2 = 9e9 * 4*1.6 e-19 *1.6 e-19 /(4*0.021*0.021)
F13 = +0.522 e -24 N
F14 = Kq1q4/(3d^4) = 9e9 * 4*1.6 e-19 *5*1.6 e-19 /(9*0.021*0.021)
F14 = +1.16 e-24 N
so
FNet on 1 = -2.1 +0.55 1.16 = -1.54*10^-24 N
-----------------------------------
Fnet on 2 = F21 + F23 + F24
F21 = 2.1 e-24 N
F23 = kq2q3/(d^2) = 9e9 *1.6 e-19 *1.6 e-19 /(0.021*0.021)
F23 = -0.522 e -24 N
F24 = Kq2q4/(2d)^2 = 9e9 *1.6 e-19 *5*1.6 e-19 /(4*0.021*0.021)
F24 = -0.653 e-24 N
s0
Fnet on 2 = 2.1-0.522-0.653 = 9.25*10^-25 N
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