The intensity of the electric field required to keep a water drop of radius 10 − 5 cm, just suspended in air when charged with one electron is approximately (g = 10newton / kg, e = 1.6 × 10 − 19 coulomb) Options: A . &nbsp 260 volt / cm B . &nbsp 260 newton / coulomb C . &nbsp 130 volt / cm D . &nbsp 130 newton / coulomb E . &nbsp If assertion is false but reason is true.

Answer: Option B : B For balance m g = e E ⇒ E = m g e Also m = 4 3 π r 3 d = 4 3 × 22 7 × ( 10 − 7 ) 3 × 1000 k g ⇒ E = 4 / 3 × 22 / 7 × ( 10 − 7 ) 3 × 1000 × 10 1.6 × 10 − 19 = 260 N / C

The intensity of the electric field required to keep a water drop of radius 10

Discussion Forum : Electric Charges Fields And Gauss Law

Question -

The intensity of the electric field required to keep a water drop of radius $10^{- 5}$ cm, just suspended in air when charged with one electron is approximately (g = 10newton / kg, e $= 1.6 \times 10^{- 19}$ coulomb)

Options:

A . 260 volt / cm

B . 260 newton / coulomb

C . 130 volt / cm

D . 130 newton / coulomb

E . If assertion is false but reason is true.

Answer: Option B
:
B

For balance $m g = e E \Rightarrow E = \frac{m g}{e}$
Also $m = \frac{4}{3} π r^{3} d = \frac{4}{3} \times \frac{22}{7} \times (10^{- 7})^{3} \times 1000 k g$
$\Rightarrow E = \frac{4 / 3 \times 22 / 7 \times (10^{- 7})^{3} \times 1000 \times 10}{1.6 \times 10^{- 19}} = 260 N / C$

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