**Operation of an Inkjet Printer**

Part A = 3.08*10^-14

**Solution Below:**

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 = 16.0 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 D = 2.50 cm, where there is a uniform vertical electric field with magnitude E = 8.10*10^{4} N/C.

Part A

^{3}, and ignore the effects of gravity.

Express your answer numerically in coulombs.

Remember that the force experienced by the drop is equal to:

F = qE

Since F = ma, we also know that:

ma = qE

and:

a = qE/m

Now we can use the acceleration to find the change in position (deflection):

x = 1/2at^{2}

x = 1/2 * (qE/m) * t^{2}

And we can use this to find q:

x = 1/2 * (qE/m) * t^{2}

2x / (t^{2}) = qE/m

q = (2 * x * m) / (E * t^{2})

Since the plates are 2.5cm long, the time will be:

t = 0.025 m / 16 m/s

t = 0.0015625

Therefore:

q = (2 * d * m) / (E * t^{2})

q = (2 * 0.00032 * (1*10^-11)) / ((8.5*10^4) * (0.0015625^2))

q = 3.08*10^-14

Make sure you converted all of the units properly when solving this problem!

8.17*10^-14 C

I’m doing this problem the numbers are different but I converted correctly and keep getting the answer 8.42×10^-14. the only numbers that differ are the velocity is 24m/s, the D is 2.4cm, and the E is 7.60×10^4 what am I doing wrong?

That is the right answer. We had accidentally mixed up the velocity and distance (D), but have fixed this and the problem uses the right numbers now. Your answer for the numbers you have should definitely be correct.