The magnitude of electric field required to just balance in air a 2× 10- 4kg liquid drop carrying a charge of 10× 10- 2 textμC is,

Question

The magnitude of electric field required to just balance in air a 2× 10- 4kg liquid drop carrying a charge of 10× 10- 2 textμC is, (A) 104NC- 1 (B) 2× 104NC- 1 (C) 4× 104NC- 1 (D) 5× 104NC- 1

Tardigrade Admin · Accepted Answer

qE=mg E=(mg/q)=(2 × 10- 4 × 10/10 × 10- 2 × 10- 6)=2× 104NC- 1

Q. The magnitude of electric field required to just balance in air a $2 \times 1 0^{- 4} k g$ liquid drop carrying a charge of $10 \times 1 0^{- 2} μC$ is,

$1 0^{4} N C^{- 1}$

$2 \times 1 0^{4} N C^{- 1}$

$4 \times 1 0^{4} N C^{- 1}$

$5 \times 1 0^{4} N C^{- 1}$

$qE = m g$
$E = \frac{m g}{q} = \frac{2 \times 1 0 ^{- 4} \times 10}{10 \times 1 0 ^{- 2} \times 1 0 ^{- 6}} = 2 \times 1 0^{4} N C^{- 1}$

Q. The magnitude of electric field required to just balance in air a 2×10−4kg liquid drop carrying a charge of 10×10−2μC is,

104NC−1

2×104NC−1

4×104NC−1

5×104NC−1