Answer:
Explanation:
Given:
Charge = <em>q</em>
Electric field strength =
weight of the droplet = <em>mg</em>
The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.
electric force on charged droplet, 
This is balanced by the weight, 
Equating the two:

Answer:
The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
Explanation:
We can find the drift speed by using the following equation:
Where:
I: is the current = 4.50 A
n: is the number of electrons
q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C
A: is the cross-sectional area = 2.20x10⁻⁶ m²
We need to find the number of electrons:
Now, we can find the drift speed:
Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
I hope it helps you!