As given that net electrostatic force on q4 = 0
Force on q4 due to other charges,
kq1q4 / x1^2 + kq2q4 / x2^2 + kq3q4 / x3^2 = 0
k*q4 (q1 / x1^2 + q2 / x2^2 + q3 / x3^2) = 0
(q1 / x1^2 + q2 / x2^2 + q3 / x3^2) = 0
(6.08*10-6 / 7.152) - (4.51*10-6 / 15.22) + (q3 / 21.62) = 0
q3 = -46.38*10-6 C
q3 = -46.38 uC
nt Chapter 21, Problem 047 Point charges of q1-6.08 uC and 2451 C are placed on...
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Chapter 18, Problem 21 (Vector Drawing) The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 but opposite signs: q2 (magnitude and direction) exerted on q1 by the other two charges. (c) If q1 had a mass of 1.50 g and it were free to move, what would be its acceleration? +8.00 HC; the other two charges have identical magnitudes, = +5.00 HC. (a) Draw the free-body diagram showing the...