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lesya [120]
3 years ago
14

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

plates if a potential difference of 126 V is applied?
Physics
1 answer:
erik [133]3 years ago
6 0

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

C = \frac{\kappa  \epsilon _{o} A }{d}

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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