A parallel-plate capacitor has capacitance C0 in vacuum. If a dielectric slab of dielectric constant kappa = 4.0 is inserted filling the gap, and the plate separation is halved (d / 2), what is the new capacitance?
Answer and explanation
Correct answer: D. 8 C0
- Step 1: Capacitance formula: C = kappa epsilon_0 A / d. Step 2. With kappa' = 4 kappa_0 and d' = d / 2: C' = 4 epsilon_0 A / (d / 2) = 8 (epsilon_0 A / d) = 8 C0. Answer: 8 C0. Why not the others: 2 C0 forgets kappa; 4 C0 forgets the halving of d; 16 C0 squares kappa.
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