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PRACTICE IT Use the worked ex negative charge while dropping through a vacuum (pressure 0) in an experimentr An electric field of magnitude 6.3s x 104 N/c points straight down. ample above to help you solve this problem. Tiny droplets of oll acquire a small (a) One particular droplet is observed to remain suspended against gravity If the mass of the droplet is 4.11 x 10-12 kg, find the charge carried by the droplet. (b) Another droplet of the same mass falls 8.3 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by the droplet. EXERCISE HINTS: GETTING STARTED IM STUCK! 2.00 x 105 N/C. Suppose a droplet of unknown mass remains suspended against gravity when E, What is the minimum mass of the droplet? (Take the +y direction to be upward.) kg Need Help? Read Submit Answer Save ProgressPractice Anoner Version 18 points Roint Charde
https weba yi Homework! O quadrant 3 where 1800 s 8 <2709 quadrant 4 where 270° S θ < 360。 PRACTICE IT Use the worked example above to help you solve this problem. Consider three point charges at the corners of a triangle, as shown in the figure above, where 1 6.37 x 10-9 C, 92-2.04 x 109 C, and 93 5.38 x 109 c. (a) Find the components of the force F23 exerted by 42 on g3 F23y (b) Find the components of the force F13 exerted by q1 on g3 13x (c) Find the resultant force on qs, in terms of components and also in terms of magnitude and direction. magnitude direction o counterclockwise from the +x-axis HINTS: GETTING STARTED I M STUCK! EXERCISE Using the same triangle, find the vector and direction. (Use the charges given in the Practice It section.) magnitude direction o counterclockwise from the+x-axis Need Holp? Roadh
Homework1 EXAMPLE 15.4 Electrified Oil GOAL Use electric forces and fields together with Newtons second law in a one-dimensional problem. 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 x 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 x 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. STRATEGY We use Newtons 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. Newtons law can then be solved for g. SOLUTION (A) Find the charge on the suspended droplet Apply Newtons second law to the (1) ma EE droplet in the vertical direction É points downward, hence Ey is negative. Set ay -0 in Equation (1) S2.93 x 1015 kg)9.80 m/s2) -4.85 x 10-19 C -5.92 x 104 N/C and solve for q (B) Find the charge on the falling droplet Use the kinematic displacement equation to find the acceleration: bulesinse by 103 m. 10.250 . -0.103 m-ha,(0.250 s)2 → α,--3.30 m/s2 and yo o: Solve Equation (1) for g and substitute m(ar + g). 10-15 kg)(-3.30 m/s,9.80m/s22 (2.93 X 5.92 × 104 N/C -3.22 x 10-19 C LEARN MORE REMARKS This example exhibits features similar to the Millikan Oil Drop experiment which determined the value of the fundamertal electric charge e. Notice that in both parts of the example, the charge is y nearly multiple of
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So must be neave 1 2 Tje charge The mass is minimun when Iq)刁minimum The ne minimum charge 수 1%)=e(charge on on a droplet So the minimum mas,s eledron e minimừn Mass = 3.27x10-rs k I S

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