A charged particle carrying $-1.6 \times 10^{-19}$ C is held in equilibrium against gravity between two horizontal plates 10 cm apart with a potential difference 4000 V across them. The mass of the particle is (take $g = 9.8$ m/s²):
A$1.6 \times 10^{-19}$ kg
B$6.53 \times 10^{-16}$ kg
C$4.08 \times 10^{-17}$ kg
D$6.53 \times 10^{-15}$ kg
Answer & Solution
Correct answer: B. $6.53 \times 10^{-16}$ kg
Field $E = V/d = 4000/0.10 = 4\times10^4$ V/m. For equilibrium electric force = weight: $|q|E = mg$ ⇒ $m = |q|E/g = (1.6\times10^{-19})(4\times10^4)/9.8 = 6.53\times10^{-16}$ kg.
Related questions
If a positive point charge is placed at the centre of a spherical conducting shell of radiThe SI unit of electric field isFor a charge distribution with volume density ρ, the total charge Q equalsThe electric flux through a closed surface enclosing a charge of +5 μC is (ε₀ = 8.85 × 10⁻Two charges +q and −q of equal magnitude are placed at the corners A and B of an equilaterThe principle of quantisation of electric charge is stated asAn electric dipole in a uniform electric field E experiencesThe electric field on the axial line of a short electric dipole at distance r is