Use the electric forces and fields together with Newton\'s second law in a one -

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Use the electric forces and fields together with Newton's second law in a one - dimensional problem. Tiny droplets of oil acquire a small negative charge while dropping through a vacuum (pressure = 0) in an experiment. An electric field of magnitude 5.92 times 104 N/C points straight down. (a) One particular droplet is observed to remain suspended against gravity. If the mass of the droplet is 2.93 times 10 - 15 kg, find the charge carried by the droplet. (b) Another droplet of the same mass falls 10.3 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by the droplet. We use Newton's second law with both gravitational and electric forces. In both parts the electric field E is pointing down, taken as the negative y - direction, as usual. In part (a) the acceleration is equal to zero. In part (b) the acceleration is uniform, so the kinematic equations yield the acceleration. Newton's law can then be solved for q. 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 times 10 - 15 kg)(9.80 m/s2)/ - 5.92 times 104 N/C = - 4.85 times 10 - 9 C Find the charge on falling droplet. Use the kinematic displacement equation to find the acceleration: Delta y = 1/2ayt2 + v0t 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 times 10 - 15 kg)( - 3.30 m/s2 + 9.80 m/s2)/ - 5.92 times 104 N/C = - 3.22 times 10 - 19 C This example exhibits feature similar to the Millikan Oil - Drop experiment which determined the values of the fundamentals electric charges e. Notice that in both parts of the example, the charge is very nearly a multiple of e. What would be the acceleration of the oil droplet in part (a) if the electric field suddenly reversed direction without changing in magnitude? (Select all that apply.) The magnitude of the acceleration would be 0 m/s2. The magnitude of the acceleration would be 9.80 m/s2. The magnitude of the acceleration would be 19.60 m/s2. The acceleration would be downward. The acceleration would be upward. Use the worked example above to help you solve this problem. Tiny droplets of oil acquire a small negative charge while dropping through a vacuum (pressure = 0) in an experiment. An electric field of magnitude 5.95 times 104 N/C points straight down. One particular droplet is observed to remain suspended against gravity. If the mass of the droplet is 3.84 times 10 - 12 kg, find the charge carried by the droplet. Another droplet of the same mass fails 7.6 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by the droplet. Suppose a droplet of unknown mass remains suspended against gravity when Ey = - 2.90 times 10 - 5 N/C. What is the minimum mass of the droplet? (Take the + y direction to be upward.)

Explanation / Answer

The acceleration will be downwards and the magnitude will be 19.6 m/s^2 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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