A parallel-plate capacitor has dimensions 4.0 cm × 5.0 cm.The plates are separated by a 1.0-mm thickness of paper (dielectric constant κ = 3.7).What is the charge that can be stored on this capacitor,when connected to a 6.0-V battery? (ε₀ = 8.85 × 10−¹² C²/N⋅m²) A) 4.8 × 10−⁹ C B) 4.8 × 10−¹¹ C C) 9.8 × 10−¹¹ C D) 3.9 × 10−¹⁰ C

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Quizplus · Accepted Answer

The capacitance C of a parallel-plate capacitor is given by C = (κε₀A)/d, where A is the area of the plates and d is the separation. Here, A = 4.0 cm × 5.0 cm = 20 × 10⁻⁴ m², d = 1.0 mm = 1.0 × 10⁻³ m, κ = 3.7, and ε₀ = 8.85 × 10⁻¹² C²/N⋅m². Substituting these values, C = (3.7 × 8.85 × 10⁻¹² × 20 × 10⁻⁴) / (1.0 × 10⁻³) = 6.54 × 10⁻¹¹ F. The charge Q stored is Q = CV = 6.54 × 10⁻¹¹ F × 6.0 V = 3.9 × 10⁻¹⁰ C.

A Parallel-Plate Capacitor Has Dimensions 4