KCET2013PhysicsElectrostatics
A small oil drop of mass 10⁻⁶ ~kg is hanging in at rest between two plates separated by 1 ~mm having a potential difference of 500 ~V . The charge on the drop is (g=10 ~ms ⁻² . )
Options
- A2 10⁻⁹ C
- B2 10⁻¹¹ C
- C2 10⁻⁶ C
- D2 10⁻⁹ C
Correct answer
B. 2 10⁻¹¹ C
Step-by-step solution
Given, the drop rests between the two plates aligned & q E=m g & or q V r =m g ( E= V r ) & q= m g r V & here, m=10⁻⁶ ~kg , g=10 ~m / s ², r=1 ~mm =10⁻³ ~m & and & Substituting all the values, we get &q= 10⁻⁶ 10 10⁻³ 500 , q=2 10⁻¹¹ C aligned