In Figure (a) three positively charged particles are fixed on an x axis Particle
ID: 1572236 • Letter: I
Question
In Figure (a) three positively charged particles are fixed on an x axis Particles B and Care so close to each other that they can be considered to be at the same distance from particle A. The net force on particle A due to particles B and C is 4.12 x 1 23 Nin the negative direction of the x axis. In Figure (b), particle B has been moved to the opposite side of A but is still at the same distance from it. The net force on A is now 2.95 x 10 24 N in the negative direction of the x axis. What is the ratio adaB? BC (a) (b) Numbe 1.397 Units the tolerance is +/-2%Explanation / Answer
in figure a)
force is given as
F = k (qb + qc) qa/r2
4.12 x 10-23 = (9 x 109) (qb + qc) qa/r2 eq-1
in figure b)
force is given as
F = k (- qb + qc) qa/r2
2.95 x 10-24 = (9 x 109) (- qb + qc) qa/r2 eq-2
dividing eq-1 by eq-2
(qb + qc)/(- qb + qc) = 13.96/1
((qb + qc) + (- qb + qc)) /((qb + qc) - ((-qb + qc))) = (13.96 + 1)/(13.96 - 1)
qc/qb = 1.15