Two metal plates (A, B) are kept horizontally with separation of ( 12 ) cm, with plate A on the top. An atomizer jet sprays oil (density 1.5 g/cm3) droplets of radius 1 mm horizontally. All oil droplets carry a charge 5 nC. The potentials VA and VB are required on plates A and B respectively in order to ensure the droplets do not descend. The values of VA and VB are . (Neglect the air resistance to the droplets and take g=10 m/s2)
- 100 V and 580 V
- 580 V and 100 V
- 60 V and 400 V
- 0 V and -200 V
Answer
(1) 100 V and 580 V
Solution
Mass of a drop: m = 43 r3 = 1500 43 (10-3)3 = 2 10-6 kg
For the drop not to descend: qE = mg
E = mgq = 2 10-6 105 10-9 = 4000 V/m
Potential difference: V = Ed = 4000 12 10-2 = 480 V
The force on the (positive) charge must be upward, so E points upward, from B to A.
∴ The lower plate B is at the higher potential: VB - VA = 480 V
Only VA = 100 V, VB = 580 V satisfies this.
Answer: (1)