Question
A dust particle of mass 4 10^ -12 mg in suspended in air under the influence of an electric field of 50 ~N / C directed vertically upwards. How many electrons were removed from the neutral dust particle? [Take, g=10 ~m / s ^2 ]
BITSAT
Physics
Electrostatics
2024
BITSAT 2024 (Memory Based Paper 3)
BITSAT Physics Question (2024) — Solution
Options
- A. 15
- B. 8
- C. 5
- D. 4
Answer
C. 5
Step-by-step solution
Mass of dust particle, m=4 10^ -12 mg aligned & =4 10^ -12 10^ -3 ~g \\ & =4 10^ -12 10^ -3 10^ -3 ~kg \\ & =4 10^ -18 ~kg aligned Electric field, E=50 ~N / C Weight of dust particle, W=m g =4 10^ -18 10=4 10^ -17 ~N Electric force experienced by dust particle, aligned & F_e=q E \\ & F_e=n e E=n 1.6 10^ -19 50 aligned where, n is the number of electrons removed from neutral dust particle. At balance condition, aligned & Electric force = Weight of dust particle \\ & n 1.6 10^ -19 50=4 10^ -17 \\ & n= 4 10^ -17 1.6 10^ -19 50 \\ & = 400 80 =5 aligned
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