Q#41 (Past Exam Paper – November 2019 Paper 12 Q10)
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?
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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