Question-25 The equivalent capacitance between plates ‘A’ and ‘B’ (A-area of each plate, d-separation between plates) (\varepsilon_0 – permittivity of free space) is
(A) \frac{A\varepsilon_0}{d}
(B) \frac{2A\varepsilon_0}{d}
(C) \frac{4A\varepsilon_0}{d}
(D) \frac{8A\varepsilon_0}{d}
Answer : (C)
Explanation:
The equivalent circuit is given below
C₂ and C₃ are connected in series, and this combination is then connected in parallel with C₁.
C₁ = \frac{K_1\left(\frac{A}{2}\right)\varepsilon_0}{d} = \frac{2A\varepsilon_0}{d} = 2C
C₂ = \frac{K_2\left(\frac{A}{2}\right)\varepsilon_0}{\left(\frac{d}{2}\right)} = \frac{3A\varepsilon_0}{d} = 3C
C₃ = \frac{K_3\left(\frac{A}{2}\right)\varepsilon_0}{\left(\frac{d}{2}\right)} = \frac{6A\varepsilon_0}{d} = 6C
C_s = \frac{C_2 C_3}{C_2 + C_3} = \frac{3C \times 6C}{3C + 6C} = 2C
C_{eq} = C_1 + C_s = 2C + 2C = 4C = \frac{4A\varepsilon_0}{d}