A water particle of mass 10-0 mg and having a charge of 1-50 - 10-6 C stays suspended in a room- What is the magnitude of electric field in the room- what is its direction-

The correct option is m = 10 mg = 10 × 10^ 2× 10^ 3 kg Q = 1.5 × 10^ 6C But QE = mg ⇒ E = mg/Q = 10 × 10^ 5× 10/1.5 × 10^ 6 = 1000/15 = 66.7 N/C Its direc ...

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Standard XII

Physics

Electric Field

A water parti...

Question

A water particle of mass 10.0 mg and having a charge of $1.50 \times 10^{- 6} C$ stays suspended in a room. What is the magnitude of electric field in the room? what is its direction?

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Solution

$m = 10 m g = 10 \times 10^{- 2} \times 10^{- 3} k g Q = 1.5 \times 10^{- 6} C$
But QE = mg
$\Rightarrow E = \frac{m g}{Q} = \frac{10 \times 10^{- 5} \times 10}{1.5 \times 10^{- 6}} = \frac{1000}{15} = 66.7 N / C$
Its direction will be upward.

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A water particle of mass 10.0 mg and having a charge of 1.50×10−6C stays suspended in a room. What is the magnitude of electric field in the room? what is its direction?

m=10mg=10×10−2×10−3kgQ=1.5×10−6C But QE = mg ⇒E=mgQ=10×10−5×101.5×10−6=100015=66.7N/C Its direction will be upward.

A water particle of mass 10.0 mg and having a charge of $1.50 \times 10^{- 6} C$ stays suspended in a room. What is the magnitude of electric field in the room? what is its direction?

$m = 10 m g = 10 \times 10^{- 2} \times 10^{- 3} k g Q = 1.5 \times 10^{- 6} C$
But QE = mg
$\Rightarrow E = \frac{m g}{Q} = \frac{10 \times 10^{- 5} \times 10}{1.5 \times 10^{- 6}} = \frac{1000}{15} = 66.7 N / C$
Its direction will be upward.