A vertical electric field of magnitude $4.9 \times 10^{5} N/C$ just prevents a water droplet of a mass $0.1 g$ from falling. The value charge on the droplet will be
(Given $g = 9.8 {m/s}^{2}$ )

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Solution

Since the droplet is at rest
$\Rightarrow$ Net force $= 0$
$\Rightarrow m g = q E$
$\Rightarrow q = \frac{m g}{E} = 2 \times 10^{- 9} C$

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A vertical electric field of magnitude 4.9×105 N/C just prevents a water droplet of a mass 0.1 g from falling. The value charge on the droplet will be (Given g=9.8 m/s2)

Since the droplet is at rest ⇒Net force =0 ⇒mg=qE ⇒q=mgE=2×10−9 C

A vertical electric field of magnitude $4.9 \times 10^{5} N/C$ just prevents a water droplet of a mass $0.1 g$ from falling. The value charge on the droplet will be
(Given $g = 9.8 {m/s}^{2}$ )

Since the droplet is at rest
$\Rightarrow$ Net force $= 0$
$\Rightarrow m g = q E$
$\Rightarrow q = \frac{m g}{E} = 2 \times 10^{- 9} C$