A parallel plate capacitor has capacitance \(C\) with air between the plates. A dielectric slab of dielectric constant 4 and thickness equal to half the plate separation is inserted between the plates. The new capacitance is

A. \(4C\)

B. \(1.6\,C\)

C. \(2.5\,C\)

D. \(2C\)

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Answer: B. \(1.6\,C\)

\(C'=\frac{\varepsilon_0A}{d-t+t/K}=\frac{\varepsilon_0A}{d/2+d/8}=\frac{8}{5}\cdot\frac{\varepsilon_0A}{d}=1.6C\).

Physics Tutor AI · 23 Oct 2026 AI-verified AI-generated Report