I vs V - Varying Intensity

A water drop ...

Question

A water drop of mass $3.2 \times 10^{- 18}$ kg and carrying a charge of $1.6 \times 10^{- 19}$ C is suspended stationary between two plates of an electric field. Given $g = 10$ $m / s^{2}$ , the intensity of the electric field required is

2 V / m

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200 V / m

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20 V / m

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2000 V / m

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Solution

The correct option is B $200 V / m$
Force in electric field
$F = q E$
and $F = m g$ (By gravitation)
$S o, m g = q E$ (for equilibrium)

$E = \frac{m g}{q}$

$= \frac{3.2 \times 10^{- 18} \times 10}{1.6 \times 10^{- 19}}$

$= 200 V / m$

Suggest Corrections

Similar questions

+ 3.2 \times 10^{- 19}

- 3.2 \times 10^{- 19}

2.4 ˙ A

4 \times 10^{5}

10^{- 5}

= 1.6 \times 10^{- 19}

10^{- 6}

ρ = 1000 k g / m^{3}; g = 9.8 m / s^{2}

0.64 \times 10^{- 14} k g

1.6 \times 10^{- 19} C

g = 9.8 m / s^{2}

+ 3.2 \times 10^{- 19}

- 3.2 \times 10^{- 19}

2.4 ˙ A

4 \times 10^{5}

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