Question

3. In an uniform electric field one oil drop of
0.002 miligram mass and 6 electronic charge
is stable in air. What will be intensity of
electric field ?
(A) 2.04 % 1010 N/C
(B) 2.04 x 108 N/C
(C) 2.04 x 106 N/C
(D) None of these​

Answer 1

answer : option (A) 2.04 × 10^10 N/C

explanation : mass of oil drop, m = 0.002 mg = 2 × 10^-9 Kg

charge on drop, q = 6e

= 6 × 1.6 × 10^-19 C

= 9.6 × 10^-19 C

at equilibrium,

electric force on charged oil drop = weight of oil drop

or, qE = mg

or, E = mg/q

= 2 × 10^-9 × 9.8 m/s²/(9.6 × 10^-19)

= 19.6 × 10^-9/9.6 × 10^-19

= (19.6/9.6) × 10^10

= 2.04 × 10^10 N/C