Question

Capacitors are used in many applications where you need to supply a short burst of energy. A \(100.0-\mu \mathrm{F}\) capacitor in an electronic flash lamp supplies an average power of \(10.0 \mathrm{kW}\) to the lamp for $2.0 \mathrm{ms}$. (a) To what potential difference must the capacitor initially be charged? (b) What is its initial charge?

Short answer

Answer: The initial potential difference is 20 V, and the initial charge is 0.002 C or 2.0 mC.

Step-by-step solution

Determine the Initial Potential Difference

First, let's find the initial potential difference (V) using the average power equation. Average power (P_avg) = 0.5 * C * V^2 / time We are given: C = 100.0 μF, P_avg = 10.0 kW, and time = 2.0 ms. To solve for V, we rearrange the formula: V^2 = (2 * P_avg * time) / C Now, plug in the given values and convert units: V^2 = (2 * (10.0 * 10^3 W) * (2.0 * 10^-3 s)) / (100.0 * 10^-6 F) V^2 = (20,000 * 0.002) / 0.0001 V^2 = 400 Take the square root of both sides: V = 20 V So, the initial potential difference is 20 V.

Determine the Initial Charge

Next, let's find the initial charge (Q) using the formula: Charge (Q) = Capacitance (C) * Potential difference (V) We are given: C = 100.0 μF and V = 20 V. Plug in the given values and convert units: Q = (100.0 * 10^-6 F) * (20 V) Q = 0.002 C So, the initial charge is 0.002 C or 2.0 mC.