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A parallel plate capacitor with plates of area 1 m² each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is: (Take ε0 = 8.85 × 10⁻¹² C²/(N m²))

1. 9.85 × 10⁻¹⁰ C
2. 8.85 × 10⁻¹⁰ C
3. 6.85 × 10⁻¹⁰ C
4. 7.85 × 10⁻¹⁰ C

Correct answer: 8.85 × 10⁻¹⁰ C

Solution

For a parallel-plate capacitor E = sigma/epsilon0 = Q/(epsilon0*A), so Q = epsilon0*E*A = 8.85e-12 * 100 * 1 = 8.85e-10 C.

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