Question 8
A charged oil drop of
mass m, with n excess electrons, is held stationary in the
uniform electric field between two horizontal plates separated by a distance d.
The voltage between
the plates is V, the elementary charge is e and
the acceleration of free fall is g.
What is the value of n ?
A eV / mgd B mgd
/ eV C meV
/ gd D gd
/ meV
Reference: Past Exam Paper – June 2015 Paper 12 Q30
Solution:
Answer:
B.
The oil drop has an excess
of electrons, so it is negatively charged.
Charge of electron = n × e
where e is the charge of
an electron and n is the excess electrons
The oil drop would be
attracted to the positive plate, which is the upper one. So, the electric force
on the electron is upwards.
The weight of the oil drop
acts downwards.
Since the oil drop is
stationary, the resultant force on it is zero.
(Upward) Electric force =
(Downward) Weight
Electric field strength (E)
× charge (n×e) = mg
E × n×e = mg
[but E = V / d]
(V/d) × n×e = mg
n = mgd / eV
A microscopic oil drop of mass 1.1x10-12 kg and charge 3 e is suspended
ReplyDeletein the gap between two parallel metal plates.
The separation between the plates is 10.68 mm.
Calculate the potential difference between the plates, in kV (i.e. V x 103), to the nearest kV.
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