The ink drops have a mass m = 1.00×1011 kg each and leave the nozzle and travel

Question

The ink drops have a mass m = 1.00×1011 kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 25.0 m/s . The drops pass through a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length D0 = 2.35 cm , where there is a uniform vertical electric field with magnitude E = 7.80×104 N/C .If a drop is to be deflected a distance d = 0.250 mm by the time it reaches the end of the deflection plate, what magnitude of charge q must be given to the drop? Assume that the density of the ink drop is 1000 kg/m3 , and ignore the effects of gravity.

Explanation / Answer

The droplets experience a lateral acceleration as they go through the electric field, and the distance deflected is d = 1/2 a t^2 where t is the time between the plates.
But t = L / v , where L is the length of the plates and v is the drop's velocity.
So now d = 1/2 a L^2 / v^2
Rearranging,
a = 2 d v^2 / L^2
But a = F / m , where F is the electric force on the drop and m is its mass.
F / m = 2 d v^2 / L^2
But F = E q , where E is the magnitude of the electric field and q is the charge on the drop.
E q / m = 2 d v^2 / L^2
or q = 2 m d v^2 / E L^2
.
q = (2)(1 x 10^-11)(2.5 x 10^-4)(25^2) / (8.5 x 10^4)(.0235^2)
q = 6.65 x 10^-14 C

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