You decide to build an electrostatic dust filter. They work based on electrostatic interactions between charged particles of dust and electric field. Your filter (shown below, not to scale) has a vertical length L = 1 m and a horizontal 1-cm distance d between the plates. A dust particle of mass m=10−9
kg and charge q=10−16
C falls with a constant vertical velocity v=1m/s
into the filter very close to the positively charged plate.
What electric field E do you need to make the particle reach the oppositely charged plate before it falls out of the filter?
To calculate the electric field E needed to make the particle reach the oppositely charged plate before it falls out of the filter, we can use the equation of motion for charged particles in an electric field:
F = qE
Where F is the force experienced by the dust particle, q is the charge of the particle, and E is the electric field.
The force experienced by the dust particle due to gravity is:
Fg = mg
Where m is the mass of the particle and g is the acceleration due to gravity (approximately 9.81 m/s^2).
Since the dust particle is falling with a constant velocity, the electric force on the particle must balance the force due to gravity:
qE = mg
Rearranging the equation to solve for the electric field E, we get:
E = mg/q
Substitute the values given:
m = 10^-9 kg
g = 9.81 m/s^2
q = 10^-16 C
E = (10^-9 kg * 9.81 m/s^2) / 10^-16 C
E = 9.81 * 10^7 C/kg
Therefore, the electric field E needed to make the particle reach the oppositely charged plate before it falls out of the filter is 9.81 * 10^7 C/kg, or 98.1 MV/m (megavolts per meter).