Question

An oil drop carrying a charge of 2 electrons has a mass of 3.2 × 10^–17 kg. It is falling freely in air with terminal speed. The electric field required to make the drop move upwards with the same speed is-

• A. 2 × 10³ V/m
• B. 4 × 10³ V/m
• C. 3 × 10³ V/m
• D. 8 × 10³ V/m

Answer 1

Answer: (A)

2 × 10³ V/m

Answer 2

q = 2e; m = 3.2 × 10^–17kg

viscous force 6phrv = mg (initially)……(i)

Put value from (i) in (ii)

qE = 2mg

E=

$\frac { 2 \times 3.2 \times 10 ^ { - 17 } \times 10 } { 2 \times 1.6 \times 10 ^ { - 19 } }$= 2 × 10³ V/ m