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 / s 2 )

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 / s 2 ) (A) 1.6 × 10-9 C (B) 2.0 × 10-9 C (C) 3.2 × 10-9 C (D) 0.5 × 10-9 C

Tardigrade Admin · Accepted Answer

Mg = qE (0.1 × 10-3)(9.8) =4.9 × 105 q (2 × 10-4/105)= q q =2 × 10-9 C

Q. A vertical electric field of magnitude $4.9 \times 1 0^{5} N / C$ just prevents a water droplet of a mass $0.1 g$ from falling. The value of charge on the droplet will be : (Given $g = 9.8 m / s^{2}$ )

$1.6 \times 1 0^{- 9} C$

$2.0 \times 1 0^{- 9} C$

$3.2 \times 1 0^{- 9} C$

$0.5 \times 1 0^{- 9} C$

$M g = qE$
$(0.1 \times 1 0^{- 3}) (9.8)$
$= 4.9 \times 1 0^{5} q$
$\frac{2 \times 1 0 ^{- 4}}{1 0 ^{5}} = q$
$q = 2 \times 1 0^{- 9} C$

Q. A vertical electric field of magnitude 4.9×105N/C just prevents a water droplet of a mass 0.1g from falling. The value of charge on the droplet will be : (Given g=9.8m/s2 )

1.6×10−9C

2.0×10−9C

3.2×10−9C

0.5×10−9C