Question

A drop of $10^{- 6}$kg water carries $10^{- 6}$C charge. What electric field should be applied to balance it’s weight (assume g = 10 m/sec²)

• A. $10V/m,$ Upward
• B. $10V/m,$ Downward
• C. 0.1 V/m Downward
• D. $0.1V/m,$ Upward

Answer 1

Answer: (B)

$10V/m,$ Downward

Answer 2

Initial separation between the charges = 0.06m

Final separation between the charges = 0.04m

Since $F \propto \frac{1}{r^{2}}$⇒ $\frac{F_{1}}{F_{2}} = \left( \frac{r_{2}}{r_{1}} \right)^{2}$⇒ $\frac{5}{F_{2}} = \left( \frac{0.04}{0.06} \right)^{2} = \frac{4}{9}$⇒ $F_{2} = 11.25N$