Question

A drop of water of mass ${10}^{-6}kg$ has an electric charge of 1000nC. What electric field will balance the drop in air -

• A. 10 V/m upwards
• B. 10 V/m downwards
• C. 0.1 V/m downwards
• D. 0.1 V/m upwards

Answer 1

The correct option is C 0.1 V/m downwards

Mass of water $={10}^{-6}kg$

$Q=1000nC$

electric field will balances the drop $\frac{1}{2}{mV}^2=qE$

or, $\frac{1}{2}\times {10}^{-6}\times V^2=1000\times E$

or, $E=\frac{1}{2}\times \frac{{10}^{-6}}{1000}=\frac{{10}^{-9}}{2}=0.1V/m$