Question

PRACTICE IT 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 5.60 x 10 N/C points straight down. 4 (a) One particular droplet is observed to remain suspended against gravity. If the mass of the droplet is 4.08 X 1032 kg, find the charge carried by the droplet. differs significantly from the correct answer. Rework your solution from the Enter a number beginning and check each step carefully. C (b) Another droplet of the same mass falls 8.2 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by the droplet. 2.06E-19 The response you submitted has the wrong sign. G HINTS: GETTING STARTED I IM STUCK!

example is the 2nd pic.

plzz help!!

Answer 1

a)

E = electric field in the region = 5.6 x 10⁴ N/C

m = mass of the droplet = 4.08 x 10^-12 kg

q = charge on the droplet = ?

the electric force on the droplet in upward direction balances the weight of the droplet in downward direction

Electric force = weight

qE = mg

q ( 5.6 x 10⁴) = (4.08 x 10^-12) (9.8)

q = 7.14 x 10^-16 C

b)

d = distance fallen = 8.2 cm = 0.082 m

v_o = initial velocity = 0 m/s

t = time of fall = 0.250 s

a = acceleration

using the kinematics equation

d = v_o t + (0.5) a t²

0.082 = (0) (0.250) + (0.5) a (0.250)²

a = 2.624 m/s²

m = mass of droplet = 4.08 x 10^-12 kg

q = charge on the droplet

F_e = electric force on the droplet in upward direction = q E

F_g = weight of droplet in downward direction = mg

force equation for the motion of the droplet is given as

F_g - F_e = ma

mg - q E = ma

( 4.08 x 10^-12 ) (9.8) - q (5.6 x 10⁴) = (4.08 x 10^-12) (2.624)

q = 5.23 x 10^-16 C