An oil drop of radius 1 ~m is held stationary under a constant electric field of 3.65 10^4 ~N / C due to some excess electrons presence on it. If the density of oil drop is 1.26 ~g / cm ^3 , then number of excess electrons on the oil drop approximately are [Take, g=10 ~ms ⁻² ]

Step-by-step solution

E=3.65 10^4 ~N / C , r=1 ~m =10⁻⁶ ~m aligned _ oil & =1.26 ~g / cm ^3 & =1.26 10^3 ~kg / m ^3 aligned Since, droplet is stationary hence weight of droplet = force due to electric field 4 3 r^3 _ oil g=q E...(i) If n be the number of excess electrons in the oil drop, then q=n e Hence, from Eq. (i), 4 3 r^3 _ oil g=n e E aligned n & = 4 r^3 _ oill g 3 e E & = 4 3.14 (10⁻⁶ )^3 1.26 10^3 10 3 1.6 10⁻¹⁹ 3.65 10^4 & =9.03 9 aligned