A small droplet of oil with mass of 1.84×10-15 kg is held suspended in a region of uniform electric field directed upward with a magnitude of 6625 N/C.
Is the excess charge on the droplet positive or negative? NEGATIVE
b.How many excess elementary charges reside on the droplet? (The elementary charge is e = 1.60×10-19 C.)
To determine the number of excess elementary charges residing on the droplet, we need to use the fact that the electric force acting on the droplet is equal to the gravitational force acting on it.
First, let's find the gravitational force on the droplet.
The formula for gravitational force is given by:
F_gravity = mass × acceleration due to gravity
mass = 1.84×10^(-15) kg
acceleration due to gravity = 9.8 m/s^2
F_gravity = (1.84×10^(-15)) × 9.8 ≈ 1.8032×10^(-14) N
The electric force acting on the droplet is given by:
F_electric = charge × electric field
charge = ?
electric field = 6625 N/C
Since the droplet is suspended, the electric force and gravitational force are equal in magnitude:
F_electric = F_gravity
So, we have:
charge × electric field = mass × acceleration due to gravity
Substituting the values:
charge × 6625 N/C = (1.84×10^(-15)) × 9.8
charge = [(1.84×10^(-15)) × 9.8] / 6625
charge ≈ 2.727×10^(-19) C
Now, to find the number of excess elementary charges, we divide the charge by the elementary charge (e):
number of excess elementary charges = charge / elementary charge
Substituting the values:
number of excess elementary charges ≈ (2.727×10^(-19)) / (1.60×10^(-19))
number of excess elementary charges ≈ 1.705
Therefore, there are approximately 1.705 excess elementary charges residing on the droplet.