Question 1
The diagram shows an oil droplet
that has become charged by gaining five electrons. The droplet remains
stationary between charged plates.
What is the magnitude and direction
of the electrostatic force on the oil droplet?
A 5.0 × 10–15 N upwards
B 5.0 × 10–15 N downwards
C 5.0 × 10–13 N upwards
D 5.0 × 10–13 N downwards
Reference: Past Exam Paper – June 2008 Paper 1 Q31
Solution 1:
Answer: C.
Consider the forces acting on the
charged oil droplet.
The weight of the oil droplets
causes a downward force.
Being negatively charged (due to the
electrons), the oil droplet is under an upward electrostatic force attracting
it to the positive plate.
Electrostatic force = EQ
where E is the electric field
strength and Q is the total charge of the droplet
Electric field strength = V / d =
5000 / (0.8×10-2)
Total charge = 5 × charge of an electron = 5 × (1.6×10-19)
Electrostatic force = EQ = (V/d) × Q
= 5.0 × 10-13 N
What does "charge of an electric" mean here?
ReplyDeletewhere you multiplied 5*charge of an electric field
it should be 'charge of an electron' instead of electric because it has gained 5 electrons
Deletethanks. it has been corrected above
Ahh, got it, many thanks
Delete