The equivalent capacitance between plates ' A ' and ' B ' ( A -area of each plate, d-separation between plates) ( ε₀ - permittivity of free space) is (a) (Aε₀)/(d) (b) (2Aε₀)/(d) (c) (4Aε₀)/(d) (d) (8Aε₀)/(d)

Answer: (c). Left half ( K₁, area A/2, gap d): C₁ = (4ε₀ (A/2))/(d) = (2ε₀ A)/(d). Right half: K₂ and K₃ layers (area A/2, gap d/2 each) in series: C₂ = (3ε₀ A)/(d), C₃ = (6ε₀ A)/(d). Series: (3× 6)/(3+6)(ε₀ A)/(d) = (2ε₀ A)/(d). Two halves in parallel: C = (2ε₀ A)/(d) + (2ε₀ A)/(d) = (4ε₀ A)/(d). So the answer is (c) (4Aε₀)/(d).

The equivalent capacitance between plates ' A ' and ' B ' ( A -area… | MHT CET 2025 (21 Apr, Shift 1)

Q26MHT CET 2025 · 21 APR · SHIFT 1Medium

Combination of Capacitors

The equivalent capacitance between plates ' A ' and ' B ' ( A -area of each plate, d-separation between plates) ( $\varepsilon_0$ - permittivity of free space) is

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Answer: (c)

Left half ( $K_1$ , area $A/2$ , gap $d$ ): $C_1 = \frac{4\varepsilon_0 (A/2)}{d} = \frac{2\varepsilon_0 A}{d}$ .
Right half: $K_2$ and $K_3$ layers (area $A/2$ , gap $d/2$ each) in series: $C_2 = \frac{3\varepsilon_0 A}{d}$ , $C_3 = \frac{6\varepsilon_0 A}{d}$ .
Series: $\frac{3\times 6}{3+6}\frac{\varepsilon_0 A}{d} = \frac{2\varepsilon_0 A}{d}$ .
Two halves in parallel: $C = \frac{2\varepsilon_0 A}{d} + \frac{2\varepsilon_0 A}{d} = \frac{4\varepsilon_0 A}{d}$ .
So the answer is (c) $\frac{4A\varepsilon_0}{d}$ .

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