A parallel plate capacitor having cross-sectional area A and separation d has air between plates. A dielectric slab with K = 4 and thickness d/2 is inserted without changing separation. The ratio of new capacitance to original capacitance is: [Diagram: Image]
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Question

A parallel plate capacitor having cross-sectional area A and separation d has air between plates. A dielectric slab with K = 4 and thickness d/2 is inserted without changing separation. The ratio of new capacitance to original capacitance is: [Diagram: Image]

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G1 Physics · Accepted Answer

Step 1: Initially C₀ = ε₀A/d. Step 2: A slab of thickness t in a gap d gives effective separation d(eff) = (d − t) + t/K. Step 3: Here t = d/2 and K = 4 ⇒ d(eff) = d/2 + d/8 = 5d/8. Step 4: C = ε₀A/(5d/8) = 8C₀/5. Hence C:C₀ = 8:5.