A small charged water drop with charge $1.5×10^{-6} C$ and mass 10 g is suspended in air in electric field E. The value of E is given by (Take acceleration due to gravity = $10 m s^{-2}$)

$6. 67 × 10^4 N/C$

The correct answer is Option (4) → $6. 67 × 10^4 N/C$

Given:

Charge, $q = 1.5 \times 10^{-6}\,C$

Mass, $m = 10\,g = 10 \times 10^{-3}\,kg = 0.01\,kg$

Acceleration due to gravity, $g = 10\,m/s^2$

For the drop to be suspended in air:

Electric force = Gravitational force

$qE = mg$

Therefore,

$E = \frac{mg}{q}$

Substitute values:

$E = \frac{0.01 \times 10}{1.5 \times 10^{-6}}$

$E = \frac{0.1}{1.5 \times 10^{-6}}$

$E = 6.67 \times 10^{4}\,V/m$

Final Answer: $E = 6.7 \times 10^{4}\,V/m$