Manipal MET2019PhysicsElectrostatics
A charged particle of mass 0.003 g is held stationary in space by placing it in a downward direction of electric field of 6 10^4 ~N / C . Then the magnitude of the charge is
Options
- A5 10⁻⁴ C
- B5 10⁻¹⁰ C
- C-18 10⁻⁶ C
- D-5 10⁻⁹ C
Correct answer
B. 5 10⁻¹⁰ C
Step-by-step solution
m=0.003 ~g =0.003 10⁻³ ~kg E=6 10^4 ~N / C The particle is stationary, so aligned & Electric force = Weight of the particle & aligned q E & =m g q & = m g E = 0.003 10⁻³ 10 6 10^4 & = 3 10⁻¹⁰ 10 6 =5 10⁻¹⁰ C aligned aligned