Two capacitors 2 µF and 4 µF connected in SERIES across 12 V. Charge on each:
A6 µC
B16 µC
C8 µC
D2.67 µC × 12
Answer & Solution
Correct answer: B. 16 µC
Series: 1/C_eq = 1/2 + 1/4 = 3/4 → C_eq = 4/3 µF. Charge Q = C_eq V = (4/3) × 12 = 16 µC. Same Q on each cap in series.
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