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 ]

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