A capacitor consists of two metal plates each 500 × 500 mm^2 and spaced 6 mm apart. The space between the metal plates is filled with a glass plate of 4 mm thickness and of paper of 2 mm thickness. The relative permittivity's of the glass and paper are 8 and 2 respectively. Neglecting the fringing effect, the capacitance will be (permittivity of the free space = 8.85 × 10^-12 F/m)
- A. 983.3 pF
- B. 1475 pF
- C. 637.5 pF
- D. 9956.25 pF
Answer: B. 1475 pF
Explanation: Capacitance C = ε0 εr A / d = (8.854×10^-12 × 4 × 0.25) / 0.006 ≈ 1475 pF.