The electric field strength in N C -1 that is required to just prevent a water drop carrying a charge 1.6 × 10-19 C from falling under gravity is (g = 9.8 ms-2, mass of water drop = 0.0016 g)

Question

The electric field strength in N C -1 that is required to just prevent a water drop carrying a charge 1.6 × 10-19 C from falling under gravity is (g = 9.8 ms-2, mass of water drop = 0.0016 g) (A) 9.8 × 1016 (B) 9.8 × 10-16 (C) 9.8× 10-13 (D) 9.8× 1013

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/773ff1a7df30ec1a29227f6b3facf7c4-.png" /> Charge on drop =1.6 × 10-19 C Mass of the drop =0.0016 g =16 × 10-4 g =16 × 10-7 kg g=9.8 m / s -2 In balance position, m g=q E where, E is electric field strength. E =(m g/q)=(16 × 10-7 × 9.8/1.6 × 10-19) =9.8 × 1013 N / C

Q. The electric field strength in $N C^{- 1}$ that is required to just prevent a water drop carrying a charge $1.6 \times 1 0^{- 19} C$ from falling under gravity is $(g = 9.8 m s^{- 2}$ , mass of water drop = $0.0016 g$ )

$9.8 \times 1 0^{16}$

$9.8 \times 1 0^{- 16}$

$9.8 \times 1 0^{- 13}$

$9.8 \times 1 0^{13}$

$9.8 \times 1 0^{- 10}$

Q. The electric field strength in NC−1 that is required to just prevent a water drop carrying a charge 1.6×10−19C from falling under gravity is (g=9.8ms−2, mass of water drop = 0.0016g)

9.8×1016

9.8×10−16

9.8×10−13

9.8×1013

9.8×10−10