Question

A drop of water of radius 10^-4 cm is freely hanging in air, in the electric field of 500 N/C. Calculate the charge on the water drop.

(please answer step by step)​

Answer 1

Answer:

Let the electric field required be E.

According to equilibrium condition,

34π(10−6)3×1000×9.8=1.6×10−19×E

⇒E≈2.566×105V/m

Solve any question of Dual Nature of Radiation And Matter with:-

Answer 2

Answer: mg = eE

$m = 4.19 \times 10^{-3}Kg$

$g = 10 m/s^2\\e = 1.6\times 10^{-19} C$

$4.19 \times 10^{-15}\times 10 = q \times 500$

$q =ne\\\\q = n\times 1.6\times 10^{-19}$

$mg =qE\\q = \frac{mg}{E}\\\\= \frac{\frac{4}{3}\times\frac{22}{7}\times{10}^{-18}\times1000\times 10}{500}\\\\q = 0.00838\times10^{-14}\\\\n=\frac{q}{e}\\n = 0.00838\times10^{-14}}/{1.6\times10^{-19}}\\n = 524$

Explanation: When a drop is suspended in the air there will be

force due to gravity, F =mg acting upward and it carries a charge with the electric field of 500 N/C there will be force due to electric field, F =eE acting downward.

At the equilibrium both the force become equal

mg =eE

m =mass

g = acceleration due to gravity

e= charge

E = electric field

first, find the mass

here we have considered the density of water approx 1000 $Kg/m^3$

$r = 10^{-4} cm = 10^{-6} m$

$mass = \frac{4}{3}\pi r^3 \rho\\ = \frac{4}{3}\times \frac{22}{7}\times 10^{-18} \times 1000.$

$m = 4.19 \times 10^{-15}Kg$