The electric field required to keep a water drop of mass m just remain suspended when charged with one electron is

  1. \( \frac {mg}{e}\)
  2. mge
  3. \(\frac{eg}{m}\)
  4. \(\frac{em}{g}\)

Answer (Detailed Solution Below)

Option 1 : \( \frac {mg}{e}\)
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Detailed Solution

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CONCEPT:

  • The electric field is the electric force experienced for a unit charge when placed in an electric field and is given by

\(​⇒ E = \frac{F}{q}\)

Where F = Force, q = charge

  • Gravitational force is the attractive force exerted by the earth on any massive object.
  • If a charged particle is placed under the influence of gravity, then for equilibrium the electrical force must be equal to the gravitational force

​⇒ FE = Fg

​⇒ qE = mg

Where m = mass, g = acceleration due to gravity, E = Electric field

EXPLANATION:

  • If a charged particle is placed under the influence of gravity, for equilibrium gravitational force must be equal to the electrical force

​​⇒ FE = Fg

​⇒ qE = mg

  • For an electron, the above equation can be written as

⇒ eE = mg

\(⇒ E = \frac{mg}{e}\)

  • Hence, option 1 is the answer
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