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Homework Statement:

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 2.0 μg dust particle is suspended in midair just above the center of the carpet.
What is the charge on the dust particle?
Relevant Equations:

F= ma
E= σ/2e0 (σ is the surface charge density and e0 is epsilon naught)
E= F/q
At first I take the uniformly distributed charge and then divide it by the area of the carpet to get the surface charge density σ
10E6 C / 8m^2 = σ = 1.25E6C/m^2
Then I divide the surface charge density by 2e0 to get the electric field strength caused by the infinite plane
1.25E6/(2(8.85E12 C^2/N.m^2 )) = 700621. N/C = E
Then I take the gravitational force on the particle
F = (2E6)(9.81m/s^2) = 1.962E5 N
Then I re arrange the formula of E=F/q to be have q on one side and then I substitute in numbers.
qE= F q= F/E
(1.962E5 N) / (700621N/C) = 2.778192 E10 C or 2.8 E10 C
10E6 C / 8m^2 = σ = 1.25E6C/m^2
Then I divide the surface charge density by 2e0 to get the electric field strength caused by the infinite plane
1.25E6/(2(8.85E12 C^2/N.m^2 )) = 700621. N/C = E
Then I take the gravitational force on the particle
F = (2E6)(9.81m/s^2) = 1.962E5 N
Then I re arrange the formula of E=F/q to be have q on one side and then I substitute in numbers.
qE= F q= F/E
(1.962E5 N) / (700621N/C) = 2.778192 E10 C or 2.8 E10 C