Question

The ink drops have a mass m = 1.00

Answer 1

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)(2.6 x 10^-4)(22^2) / (7.5 x 10^4)(.0245^2)
q = Answer in C