Question

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).

पर्याय

• 1.6 × 10^-9 C

• 2.0 × 10^-9 C

• 3.2 × 10^-9 C

• 0.5 × 10^-9 C

Answer 1

2.0 × 10^-9 C

Explanation:

As water droplet is at rest

So, `vec"F"_"net"= 0`

⇒ mg = qE ⇒ q = `"mg"/"E"`

⇒ q = `(0.1xx10^-3xx9.8)/(4.9xx10^5)`

⇒ q = 2 × 10^-9 C