Question

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 1

Correct Answer - Option 1 : \( \frac {mg}{e}\)

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

​⇒ F_E = F_g

​⇒ 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

​​⇒ F_E = F_g

​⇒ qE = mg

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

⇒ eE = mg

\(⇒ E = \frac{mg}{e}\)

• Hence, option 1 is the answer