Question

Two point charges q1q1 and q2q2 are held in place 4.50 cmcm apart. Another point charge -2.35 μCμC of mass 4.00 gg is initially located 3.00 cmcm from each of these charges (the figure(Figure 1)) and released from rest. You observe that the initial acceleration of -2.35 μCμC is 354 m/s2m/s2 upward, parallel to the line connecting the two point charges.

Find q1 and q209 -3.00 cm K 4.50 cm K 3.00 cm.

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А 4 Let I, is tve & q, is ove 015-8- o = -2.3540 m = 4.00g = 4 x 10 kg ---.c 4.50cm - 3cm - a = 354m/s2 2 0.75 AC= AB = 2.25cx 10 11 For = 2х3х10х2.3sy of 1 Иyo: 15 = 3s 2 3 х (оч F = ma , = чинь ta, a = 35ч и/s? ma = 3.5 25 x 16 1, ( give y) чхох 3ѕ

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