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Answer #1

1 q1q2 Fc = 41972

for 1.6uC and 7uC

r = 0.53m

\frac{1}{4\pi \epsilon } = 1/ 4\pi * 8.8*10^{-12} = 9.04 * 10^9 (let this be K)

Force = K (1.6*7)/ .53^2 = 39.87K

for pair -4 and 7

F = k (-4)*7/ .53^2 = -99.67K

Net force = 86.91 * 9.04 * 10^9 N = 8.017 N

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j Three charged particles are located at the corners of an equilateral triangle as shown in...
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