Question

# An electric field can just support a water droplet $$1.0\times 10^{-6}$$m in diameter carrying one electron charge. The magnitude of electric field strength is

A
3.21×104V/m
B
2.31×104V/m
C
1.32×104V/m
D
6.42×104V/m

Solution

## The correct option is A $$3.21\times 10^{4}V/m$$Answer is A.Force in electric field$$F = qE$$Mass of water droplet$$= f_wV$$$$= f_w \dfrac{4}{3}\pi r^3$$$$= f_w \dfrac{4}{3}\pi (\dfrac{d}{2})^3$$$$= 1000\times \dfrac{4}{3}\times \pi \times (\dfrac{{10}^{-6}}{2})^3$$$$= 1000\times \dfrac{4}{3} \times \pi \times\dfrac{1}{8} \times 10^{-18}$$$$= 5.23 \times 10^{-16}kg$$So, gravitational force $$= mg$$$$= 5.23 \times 10^{-16} \times 10$$$$= 5.23 \times 10^{-15}N$$Now, for equilibrium$$qE =$$ gravitational force$$E = \dfrac{5.23 \times 10^{-15}}{1.6 \times 10^{-19}}$$     ($$\because electron charge = 1.6\times 10^{-19}C$$)$$= 3.21 \times 10^4 \dfrac{V}{m}$$Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More