Answer:
Electric field intensity is the force experienced by a test charge q in a electric field E.
Answer:
q = 4.87 X 10^ -14 C
Explanation:
As d=0.350 mm
The ink drop will be accelerated by the electric field between the plates:
a = F/m
d = a(D0 / v)^2 / 2 ...... 1
a = qE/m ............... 2
Substituting 2 into 1:
d = (qE/m)(D0 / v)^2 / 2
q = 2mdv^2 / [E(D0)^2]
q = 2(1.00e-11 kg)(3.50e-4 m)(15.0 m/s)^2 / [(7.70e4 N/C)(2.05e-2 m)^2]
q = 4.87e-14 C
