The voltage between the two plates inside of the air cleaner is 500 V. A soot particle whose mass is 10-12 kg and carries a charge of -1.0 x 10-11 C of charge enters the air cleaner. What will be the velocity of the soot particle as it moves from the negative to the positive plate?

To determine the velocity of the soot particle as it moves between the plates, we can use the principles of electrostatics and energy conservation.

The electric potential difference between the two plates is given as 500 V, and the soot particle carries a charge of -1.0 x 10^-11 C.

1. First, let's calculate the electric potential energy of the soot particle at each plate using the formula:
Electric potential energy = charge * electric potential difference
Electric potential energy at the positive plate = (-1.0 x 10^-11 C) * 500 V = -5.0 x 10^-9 J
Electric potential energy at the negative plate = (-1.0 x 10^-11 C) * 0 V = 0 J

2. Next, let's calculate the kinetic energy of the soot particle at the positive plate using the conservation of energy principle. The total mechanical energy (sum of kinetic and potential energy) remains constant in an isolated system.
Initial mechanical energy at negative plate = Electric potential energy at negative plate = 0 J
Final mechanical energy at positive plate = Kinetic energy at positive plate + Electric potential energy at positive plate

Since the soot particle starts from rest at the negative plate, its initial kinetic energy is 0 J.
Final mechanical energy = Final kinetic energy + Electric potential energy at positive plate
Final mechanical energy = 0 J + (-5.0 x 10^-9 J) = -5.0 x 10^-9 J

3. Now, let's equate the initial and final mechanical energies to find the kinetic energy of the soot particle at the positive plate:
Initial mechanical energy = Final mechanical energy
0 J = -5.0 x 10^-9 J + (1/2)mv^2, where v is the velocity of the soot particle

Solving for v^2, we get:
5.0 x 10^-9 J = (1/2)(10^-12 kg)v^2
v^2 = (2 * 5.0 x 10^-9 J) / (10^-12 kg)
v^2 = 1.0 x 10^-6 m^2/s^2

4. Finally, taking the square root of both sides to solve for v, we get:
v ≈ 1.0 x 10^-3 m/s

Therefore, the velocity of the soot particle as it moves from the negative to the positive plate is approximately 1.0 x 10^-3 m/s.