Find the charge on the suspended droplet. Apply Newton\'s second law to the drop

Question

Find the charge on the suspended droplet. Apply Newton's second law to the droplet in the vertical direction. May = Fy = - mg + Eyq E points downward, hence Ey is negative. Set ay = 0 in Equation(1) and solve for q. q = mg/Ey = (2.93 * 10 - 15 kg)(9.80 m/s2)/ - 5.92 * 104 N/C = - 4.85 * 10 - 19 C Find the charge on the falling droplet. Use the kinematic displacement equation to find the acceleration: Delta1/2ayt2 + vot Substitute Delta y = - 0.103 m, t = 0.250 s, and v0 = 0: - 0.103 m + 1/2ay(0.250 s)2 rightarrow ay = - 3.30 m/s2 Solve equation(1) for q and substitute. Q = m(ay + g)/Ey = (2.93*10 - 15 kg)( - 3.30 m/s2 + 9.80 m/s2)/ - 5.92*104 N/C = - 3.22*10 - 19 C This example exhibits features similar to the Millikan Oil - Drop experiment which determined the value of the fundamental electric charge e. Notice that in both parts of the example, the charge is very nearly a multiple of e.

Explanation / Answer

Practice: a) for equilibrium , mg=Eq q=mg/E = 3.84e-12 * 9.8 / 5.95e4 = 6.32*10^-16 C b) using the equation in the example, a=2S/t^2 = 2.43 m/s^2 so m(g-a)=Eq q = 4.756*10^-16 C Exercise: Minimum charge on droplet, q = 1.6*10^-19 C So mass = qE /g = 1.6*10^-19 * -2.90*10^5 / 9.8 = 4.73*10^-15 kg

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