Question Detail
Question
A 10 MΩ resistor is connected in series with a 1.0 µF capacitor, which is initially charged with 5.0 µC. If the battery is disconnected, approximately how long will it take for the charge on the capacitor to drop to 0.50 µC?
Correct Answer
Option C is the correct answer.
Detailed Explanation
First, calculate the time constant τ = RC = (10 × 10⁶ Ω) × (1.0 × 10⁻⁶ F) = 10 seconds. For discharging, the charge is given by q(t) = Q₀ e⁻ᵗ/ᴿᶜ. We have Q₀ = 5.0 µC and we want q(t) = 0.50 µC. So, 0.50 = 5.0 e⁻ᵗ/¹⁰. Dividing by 5.0 gives 0.1 = e⁻ᵗ/¹⁰. Taking the natural logarithm of both sides: ln(0.1) = -t/10. Since ln(0.1) ≈ -2.3026, we get -2.3026 = -t/10, which means t ≈ 23.026 seconds. This is approximately 23 seconds, as stated on page 34.
Hint
Use the discharging equation q(t) = Q₀ * e^(-t/RC) and solve for t. Remember that ln(x) is the inverse of e^x.