Question

hree identical charged particles with charges, qi-n-ga = 20.0 μC, and masses m = 0.5 grams lie on a straight line as shown. They are released from rest at their positions. What will be the speed of gn when it is very far away from the other tw particles? 3 points a1 25 cm 25 cm

Answer 1

we have a_1 = a_2 = a_3 = 20mu c = 20 times 10^-6 c masses m_1 = m_2 = m_3 = 0.5g = 0.5 times 10^-3 g Ab = 25 times 10^-2 = 0.25m AC = (25 + 25)cm = 0.5m potential at point A V_A = K[a _2 /AB + a_3/AB] suppose at for distance V = 0 so change for pointentail energy Delta U = U_f - U_i = q_1 (0) - q(V_A) = - Kq_1[q_2/AB + q_3/AB] = -9 times 10^9 times 20 times 106-6 [1/0.25 + 1/0.5] = -36 times 106-1 [4 + 2] = -216 times 10^-1 = -21.6J Delta U = -21.6J initially charge q_1 is at rest so K_1 = 1/2 mV^2_1 = 0J let at a for distance speed of charge q be V, so kinetic energy Kf = 1/2mV^2 change m kinetic energy = Delta K = Kf = K so Delta K = 1/2 mv^2 Now total mechanical energy is conserved that is Delta Sigma = Delta U + Delta K = 0 so - 21.6 + 1/2 mv^2 = 0 materialimplication V^2 = 21.6 times 2/m = 21.6 times 2/0.5 times 10^4 materialimplication V^2 = 86.4 times 10^3 = 8.64 times 10^4 or V = 2.993 times 10^2 m/s or V = 293.93 m/s

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