Question 41
A positively charged oil droplet falls in air in a
uniform electric field that is vertically upwards. The droplet has a constant
terminal speed v0 and
the electric field strength is E.
The magnitude of the force due to air resistance acting
on the droplet is proportional to the speed of the droplet.
Which graph
shows the variation with E of v0?
Reference: Past Exam Paper – November 2019 Paper 12 Q10
Solution:
Answer:
A.
Since the oil droplet is
falling with a constant speed, it has no acceleration.
The resultant force on the
oil droplet is zero.
Sum of downward forces =
Sum of upward forces
The weight of the droplet
acts downwards.
Weight = mg
Since the droplet is
falling (moving downwards), the air resistance is upwards. This force is
proportional to the speed of the droplet.
FR = kv0 where k is a constant
The electric field is
vertically upwards. Since the direction of an electric field gives the
direction of the force on a positive charge, the force on the
positively-charged oil droplet is upwards.
Electric field strength E
= F / q
Electric force F = Eq
Sum of downward forces =
Sum of upward forces
Weight = Electric force +
Force due to air resistance
mg = Eq + kv0
A graph of speed against
field strength is plotted. So, v0 the subject of formula.
Eq + kv0 = mg
v0 = mg/k – Eq/k
Comparing with the equation
of a straight line (y = mx + c) with v0 on the y-axis and E on the
x-axis,
Gradient = - q/k
Thus, the graph is a straight
line with negative constant gradient.
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