Question

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?

Answer 1

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)