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

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