A parallel-plate capacitor has circular plates of 7.10 cm radius and 1.26 mm separation. What charge (in nC) will appear on the
1 answer:
Answer:
13.98 nC
Explanation:
Capacitance depends upon the area of the plates and their distance of separation.
Radius = r = 0.071 m
separation = d = 0.00126 m
here κ = 1 and ε₀ = 8.85 ₓ 10⁻¹² SI units , for free space.
Area = A = π r² = 0.0158 m²
C = [( 8.85 ₓ 10⁻¹² ) ( 0.0158) ]÷ (0.00126) = 1.11 x 10⁻¹⁰ F
Charge = Q = C V = ( 1.11 x 10⁻¹⁰ F )(126) = 13.98 nC
= 14 nC ( rounded to two significant digits)
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