Question

$1\times {10}^{-6}C$ charge is on a drop of water having mass ${10}^{-6}kg$ . What electric field should be applied on the drop so that it is in the balanced condition with its weight?

• A. $10$ V/m upward
• B. $10$ V /m downward
• C. $0.1$ V/m downward
• D. $0.1$ V/m upward

Answer 1

The correct option is A $10$ V/m upward

Electrostatic force $=$ Weight of the drop

$⟹Eq=mg$

Given,

$q={10}^{-6}C$

$m={10}^{-6}kg$

$g=10m/s^2$

$E=?$

Rearranging, $E=\frac{mg}{q}=\frac{{10}^{-6}\times 10}{{10}^{-6}}=10V/mupwards$