Two charges q 1 = 4 Solution Ideally, the net force would be minimized when the

Question

Two charges q1 = 4

Explanation / Answer

Ideally, the net force would be minimized when the force from each charge is equal in magnitude and opposite in direction, so F1 = -F2 where F = k (q3) (qx) / r^2 where r1= r and r2 = (19-r) F1 + F2 = 0 ---> (k*q3) [(q1/r^2)+(q2/(19-r)^2] = 0 k * q3 is not equal to zero, so the part in square brackets must be. Multiplying both sides by r^2 * (19-r)^2 : 0 = q1(19-r)^2 + q2(r^2) simplify and rearrange: 361(q1)-38(q1)(r)+(q1)(r^2) + q2 (r^2) = 361(q1)-38(q1)(r)+(q1+q2)(r^2) =0 0 = 1444 - 152r - 44r^2 Using the quadratic equation: r (cm) = {152 +/- sqrt[152^2 - 4(-44)(1444)]}/2(-44) r (cm) = (152 +/- 526.54)/(-88) = 678.54/(-88) or -374.54/(-88) = -7.71 or 4.26 The first term doesn't help because it is not between the two points, so it must be the second Remembering r is the distance from the first point, the charge should be placed approximately 4.26 cm from q1

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