If the voltage between the two plates of an electric air cleaner is 500 V, how fast would a 10-12 kg soot particle with -1 x 10-11 C of charge on it be moving if it went from the negative plate to the positive?
The electrical potential energy loss,
q*V = 5*10^-9 J ,
would equal the kinetic energy gain when it hits the opposite plate,
(1/2) M V^2
V = sqrt(2 q V/M) = 100 m/s
To find the velocity of the soot particle, we can use the principles of electrostatics and the concept of electric potential energy. The key equations we will use are:
1. Electric Potential Energy (PE) = Charge (q) x Voltage (V)
2. Kinetic Energy (KE) = 1/2 x Mass (m) x Velocity squared (v^2)
Given:
- Voltage (V) = 500 V
- Charge (q) = -1 x 10^-11 C
- Mass (m) = 10^-12 kg
Step 1: Calculate the electric potential energy (PE)
PE = q x V
PE = (-1 x 10^-11 C) x (500 V)
PE = -5 x 10^-9 J
Step 2: Calculate the kinetic energy (KE)
Since the particle starts from rest, all of the potential energy is converted into kinetic energy.
PE = KE
KE = -5 x 10^-9 J
Step 3: Solve for velocity (v)
KE = 1/2 x m x v^2
-5 x 10^-9 J = 1/2 x (10^-12 kg) x v^2
-5 x 10^-9 J = 5 x 10^-13 kg x v^2
Divide both sides by the mass:
v^2 = (-5 x 10^-9 J) / (5 x 10^-13 kg)
v^2 = -10^4 m^2/s^2
Take the square root of both sides to find the velocity:
v = √(-10^4) m/s
However, this result is imaginary, which means that the particle cannot travel from the negative plate to the positive plate. There might be an error in the given values or the setup.
Please verify the given information or provide additional details if available.