First slide

Electric field

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
$(g = 10 newton / kg, e = 1 .6 \times 10^{- 19} coulomb)$

Easy

A

$260 volt / cm$
B

$260 newton / coulomb$
C

$130 volt / cm$
D

$130 newton / coulomb$

Solution

For balance $mg = eE \Rightarrow E = \frac{mg}{e}$
Also $m = \frac{4}{3} {πr}^{3} d = \frac{4}{3} \times \frac{22}{7} \times {(10^{- 7})}^{3} \times 1000 kg$
$\Rightarrow E = \frac{4 / 3 \times 22 / 7 \times {(10^{- 7})}^{3} \times 1000 \times 10}{1 .6 \times 10^{- 19}} = 260 N / C$

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