JEE Main 2022 · 24 June · morning · Single correct · +4 / −1
A vertical electric field of magnitude 4.9 × 10⁵ N/C just prevents a water droplet of a mass 0.1 g from falling. The value of charge on the droplet will be :
(Given : g = 9.8 m/s²)
- (a) 1.6 × 10⁻⁹ C
- (b) 2.0 × 10⁻⁹ C
- (c) 3.2 × 10⁻⁹ C
- (d) 0.5 × 10⁻⁹ C
Answer and solution
Answer (b) 2.0 × 10⁻⁹ C
Solution
Physics sir · 4 steps
The droplet just hangs there, so the electric force upward exactly equals its weight: set qE = mg.
- Balance the forces. The field just prevents the droplet from falling, so the net force on it is zero. The electric force qE acts upward and the weight mg acts downward.
qE = mg ⇒ q = mg/E
- Convert the mass. m = 0.1 g = 1 × 10⁻⁴ kg, so the weight is mg = 1 × 10⁻⁴ × 9.8 = 9.8 × 10⁻⁴ N.
- Put in numbers. q = (9.8 × 10⁻⁴)/(4.9 × 10⁵) = 2.0 × 10⁻⁹ C.
- So. The charge on the droplet is 2.0 × 10⁻⁹ C, which is option (b).
The idea to keep. Whenever a charged body floats or stays balanced in a field, net force is zero, so qE = mg. Convert grams to kilograms before you divide.
Answer from the NTA final answer key. Solution by PrepWiser.