A drop of liquid has a mass 10^-14 kg charge 2e . calculate the magnitude of the electric field in vertically upward direction so the droplet is in equilibrium near the surface of earth.
Answers
It is given that a drop of liquid has a mass 10¯¹⁴ kg charge 2e¯ .
we have to calculate the magnitude of the electric field in vertically upward direction so the droplet is in equilibrium near the surface of earth.
at equilibrium,
weight of droplet = electric force acting on droplet
⇒mg = Eq
here m = 10¯¹⁴ kg , q = 2e¯ = 2 × 1.6 × 10^-19 C = 3.2 × 10^-19 C and g = 9.8 m/s²
so, 10¯¹⁴ kg × 9.8 m/s² = E × 3.2 × 10^-19 C
⇒E = 9.8 × 10¯¹⁴/(3.2 × 10^-19 )
= 3.0625 × 10^5 N/C
therefore, electric field in upward direction so that droplet is in equilibrium near the surface of earth.
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