A capacitor acquires a charge of 0.003 C when connected across a 30-V battery. Its capacitance is _____.
1 answer:
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Answer:
10.6 mA
Explanation:
t = time interval = 1.00 s
q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C
n₁ = number of Na⁺ ions = 2.68 x 10¹⁶
q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C
n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶
q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C
i₁ = Current due to Na⁺ ions = =
= 0.004288 A
i₂ = Current due to Cl⁻ ions = =
= 0.006272 A
Current passing between the electrodes is given as
i = i₁ + i₂
i = 0.004288 + 0.006272
i = 0.01056 A
i = 10.6 x 10⁻³ A
i = 10.6 mA