Question

Another droplet of the same mass falls 8.4 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by the droplet.

Answer 1

Answer:

The charge carried by the droplet is $1.330\times10^{-19}\ C$

Explanation:

Given that,

Distance =8.4 cm

Time = 0.250 s

Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude $5.92\times10^4\ N/C$ points straight down and if the mass of the droplet is $2.93\times10^{-15} kg$

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2$

$a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}$

$a=2.688\ m/s^2$

We need to calculate the charge carried by the droplet

Using formula of electric filed

$E=\dfrac{F}{q}$

$q=\dfrac{ma}{E}$

Put the value into the formula

$q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}$

$q=1.330\times10^{-19}\ C$

Hence, The charge carried by the droplet is $1.330\times10^{-19}\ C$