Question

The magnitude of electric field intensity \( E \) is such that, an electron placed in it would experience an electrical force equal to its weight. Then the value of \( E \) is:

• A. \( mge \)
• B. \( \frac{mg}{e} \)
• C. \( \frac{e}{mg} \)
• D. \( \frac{e^2}{mg} \)

Hint:

When the electrical force on a charged particle equals its weight, the electric field intensity can be calculated using \( E = \frac{mg}{e} \).

Answer 1

The Correct Option is B

The force on the electron due to the electric field is \( F = eE \). The force due to gravity is \( mg \). Equating these forces: \[ eE = mg \] Solving for \( E \): \[ E = \frac{mg}{e} \] Thus, the electric field intensity is \( \frac{mg}{e} \).