Question : Two horizontal plates, separated by 1 cm, are arranged one above the other. A particle of mass 5 mg and charge 2 nC is released in air between the plates. The potential difference that should be applied to the plates so that the particle remain suspended between them is :
(A) 250 V
(B) 200 V
(C) 100 V
(D) 50 V
Doubt by Japjee
Solution :
l = 1 cm = 1×10-2 m
m = 5 mg = 5×10-6 kg
q = 2 nC = 2×10-9 C
We know
F=mg
qE=mg
q[V/l]=mg
V=mgl/q
V= [5×10-6×10×1×10-2]/2×10-9 [∵g=10 m/s²]
V= [5/2]×10-7+9
V=2.5×102
V=250 Volts
F=mg
qE=mg
q[V/l]=mg
V=mgl/q
V= [5×10-6×10×1×10-2]/2×10-9 [∵g=10 m/s²]
V= [5/2]×10-7+9
V=2.5×102
V=250 Volts
Hence, (A) 250 V, would be the correct option.
