JEE Main2004PhysicsElectrostaticsActual
A charged oil drop is suspended in a uniform field of 3 10^4 ~V / m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge =9.9 10⁻¹⁵ ~kg and g =10 ~m / s ^2 )
Options
- A3.3 10⁻¹⁸ C
- B3.2 10⁻¹⁸ C
- C1.6 10⁻¹⁸ C
- D4.8 10⁻¹⁸ C
Correct answer
A. 3.3 10⁻¹⁸ C
Step-by-step solution
Since ball is hanging in equilibrium, force by gravity is balanced by electric force. aligned & qE = mg & q = m g E & 9.9 10⁻¹⁵ 10 3 10^4 & q =3.3 10⁻¹⁸ C aligned