The ink drops have a mass = 1.00×10-11 each and leave the nozzle and travel hori

Question

The ink drops have a mass = 1.00×10-11 each and leave the nozzle and travel horizontally toward the paper at velocity = 17.0 . The drops pass through a charging unit that gives each drop a positive charge by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length = 2.05 , where there is a uniform vertical electric field with magnitude = 8.40×104 .
Part A
If a drop is to be deflected a distance = 0.290 by the time it reaches the end of the deflection plate, what magnitude of charge must be given to the drop? Assume that the density of the ink drop is 1000 , and ignore the effects of gravity.
Express your answer numerically in coulombs.

Explanation / Answer

The timeit takes for the drops to reach the end of plates = d / v = 0.0205meters /17meters per sec = 1.206 E-3 sec So the acceleration needed for 0.29 mm deflection is = 2S/t^2 =(2*0.29e-3 / (1.206 E-3)^2) = 398.8 m/s^2 Force = m * a = 1.0 E-11 kg * 398.8 m/s^2 =3.988 E-9 N Electric force = Electric field *charge = 3.988 E-9 N / 8.4 x 10^4 N/C = 4.748 E-14 C

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