Question

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

Answer 1

Answer:

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Answer 2

The intensity of electric field is given by the equation:

$E =\frac{mg}{e}$         EQ (1)

Where, $m=\frac{4}{3}$π$r^{3} d$        ( r = radius of water droplet , d = density of water)

              g = acceleration due to gravity

               e = charge of coulomb

 Given ,  radius of water droplet (r) = $10^{-5} cm$

                                                            = $10^{-7} m$

let g = 10 $\frac{m}{s^{2} }$

e = 1.6 ×$10^{-19}\ C$

Substituting all the values in  EQ(1), we get:

              E = $\frac{\frac{4}{3} X\ 3.14\ X(10^{-7} )^{3}\ X 1000 X\ 10 }{1.6\ X\ 10^{-19} }$

Solving, we get:

E = 260 newton /coulomb

Therefore, The intensity of the electric field required is 260 N/C.

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