The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 23.0 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 = 1.80 cm , where there is a uniform vertical electric field with magnitude e = 8.20×10^4 n/c if a drop is to be deflected a distance d = 0.290 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/m^3

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time =

or

Time =

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a =

Now from the Newton's equation of motion

where,  

d is the distance

u is the initial speed  

a is the acceleration

t is the time

or

or

q = 9.98 × 10⁻⁹ C

explanation:

newton's third law, for every action there is an equal and opposite reaction.

gravity

explanation: