A vertical electric field of magnitude 4 . 9 × 10 5   N   C - 1 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   s - 2 )

Question

A vertical electric field of magnitude 4 . 9 &#215; 10 5 &#160; N &#160; C - 1 just prevents a water droplet of a mass 0 . 1 &#160; g from falling. The value of charge on the droplet will be : (Given g = 9 . 8 &#160; m &#160; s - 2 )

Aicharya · Accepted Answer

To balance gravitational force, electric force on the liquid drop should act in upward direction.

For translational equilibrium along vertical direction,

$q E = m g$

$\Rightarrow q = \frac{m g}{E} = \frac{0.1 \times 10^{- 3} \times 9.8}{4.9 \times 10^{5}} = 2.0 \times 10^{- 9} C$

A vertical electric field of magnitude 4 . 9 × 10 5   N   C - 1 just prevents a water droplet…

Solution

A vertical electric field of magnitude 4 . 9 &#215; 10 5 &#160; N &#160; C - 1 just prevents a water droplet…

Solution

A vertical electric field of magnitude 4 . 9 × 10 5 N C - 1 just prevents a water droplet…