Use the worked example above to help you solve this problem. Tiny droplets of oil acquire a smallnegative charge while dropping through a vacuum (pressure = 0) in an experiment. An electric field ofmagnitude 5.75 x 104 N/C points straight down.(a) One particular droplet is observed to remain suspended against gravity. If the mass of thedroplet is 3.48 x 10 12 kg, find the charge carried by the droplet.C(b) Another droplet of the same mass falls 7.4 cm from rest in 0.250 s, again moving through avacuum. Find the charge carried by the droplet.EXERCISEHINTS: GETTING STARTED| I'M STUCK!Suppose a droplet of unknown mass remains suspended against gravity when E = -2.10 × 105 N/C.What is the minimum mass of the droplet? (Take the +y direction to be upward.)kgNeed Help?Read It SOLUTION(A) Find the charge on the suspended droplet.Apply Newton's second law to the(1) ma, = EF, = -mg + E_qdroplet in the vertical direction.mg (2.93 x 10-15 kg)(9.80 m/s²)EyEpoints downward, hence E, is-4.85 x 10-19 cnegative. Set ay= 0 in Equation (1)-5.92 x 104 N/Cand solve for q.(B) Find the charge on the falling droplet.Use the kinematic displacementAy = ½ar² + vo?equation to find the acceleration:Substitute Ay= -0.103 m, t = 0.250 s,-0.103 m = ½a, (0.250 s)² → a, = -3.30 m/s²andVo= 0:Solve Equation (1) for q andm(a, +g) (2.93 x 10-15 kg)(-3.30 m/s2 + 9.80 m/s2)substitute.E,-5.92 x 104 N/C-3.22 x 10-19 cC

Given, Electric field ,Ey =-2.10×105 N/C Acceleration due to gravity , g = 9.8 m/s2 Minimum charge…

Answered: Suppose a droplet of unknown mass…

Suppose a droplet of unknown mass remains suspended against gravity when E. = -2.10 × 10 N/C. What is the minimum mass of the droplet? (Take the +y direction to be upward.) kg

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Relativistic Energy and Momentum

According to classical physics, time flow is the same for every observer. But according to einstein's theory, time flow is different for different observers. The idea that time flow is not the same for all may sound strange because we may not experience this effect in everyday life. This effect is only observed when an object moves with some velocity comparable to the velocity of light. This velocity is termed relativistic velocity.

Question

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 5.75 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 3.48 x 10 12 kg, find the charge carried by the droplet.
C
(b) Another droplet of the same mass falls 7.4 cm from rest in 0.250 s, again moving through a
vacuum. Find the charge carried by the droplet.
EXERCISE
HINTS: GETTING STARTED
| I'M STUCK!
Suppose a droplet of unknown mass remains suspended against gravity when E = -2.10 × 105 N/C.
What is the minimum mass of the droplet? (Take the +y direction to be upward.)
kg
Need Help?
Read It

SOLUTION
(A) Find the charge on the suspended droplet.
Apply Newton's second law to the
(1) ma, = EF, = -mg + E_q
droplet in the vertical direction.
mg (2.93 x 10-15 kg)(9.80 m/s²)
Ey
E
points downward, hence E, is
-4.85 x 10-19 c
negative. Set ay
= 0 in Equation (1)
-5.92 x 104 N/C
and solve for q.
(B) Find the charge on the falling droplet.
Use the kinematic displacement
Ay = ½ar² + vo?
equation to find the acceleration:
Substitute Ay
= -0.103 m, t = 0.250 s,
-0.103 m = ½a, (0.250 s)² → a, = -3.30 m/s²
and
Vo
= 0:
Solve Equation (1) for q and
m(a, +g) (2.93 x 10-15 kg)(-3.30 m/s2 + 9.80 m/s2)
substitute.
E,
-5.92 x 104 N/C
-3.22 x 10-19 cC

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