The current that charges a capacitor transfers energy that is stored in the capacitor's electric field. Consider a 2.0 μF capacitor, initially uncharged, that is storing energy at a constant 200 W rate. What is the capacitor voltage 2.0 μs after charging begins?

Verified step by step guidance

1. Start by recalling the formula for the power stored in a capacitor: \( P = \frac{1}{2} C \frac{dV^2}{dt} \), where \( P \) is the power, \( C \) is the capacitance, and \( \frac{dV^2}{dt} \) is the rate of change of the square of the voltage.
2. Rearrange the formula to solve for \( \frac{dV^2}{dt} \): \( \frac{dV^2}{dt} = \frac{2P}{C} \). Substitute \( P = 200 \ \text{W} \) and \( C = 2.0 \ \mu\text{F} = 2.0 \times 10^{-6} \ \text{F} \) into the equation.
3. Integrate \( \frac{dV^2}{dt} \) with respect to time to find \( V^2 \): \( V^2 = \int \frac{2P}{C} dt = \frac{2P}{C} t + V_0^2 \). Since the capacitor is initially uncharged, \( V_0 = 0 \), so \( V^2 = \frac{2P}{C} t \).
4. Substitute \( t = 2.0 \ \mu\text{s} = 2.0 \times 10^{-6} \ \text{s} \), \( P = 200 \ \text{W} \), and \( C = 2.0 \times 10^{-6} \ \text{F} \) into the equation \( V^2 = \frac{2P}{C} t \) to calculate \( V^2 \).
5. Finally, take the square root of \( V^2 \) to find the voltage \( V \): \( V = \sqrt{\frac{2P}{C} t} \). This gives the voltage across the capacitor after 2.0 \( \mu\text{s} \).

Key Concepts

Capacitance

Capacitance is the ability of a capacitor to store electric charge per unit voltage. It is measured in farads (F) and is defined as the ratio of the charge (Q) stored on one plate of the capacitor to the voltage (V) across the plates, expressed as C = Q/V. In this question, the 2.0 μF capacitor indicates its capacitance, which is crucial for determining how much charge it can store as it charges.

Energy Stored in a Capacitor

The energy (U) stored in a capacitor is given by the formula U = 1/2 CV², where C is the capacitance and V is the voltage across the capacitor. This relationship shows that the energy increases with the square of the voltage, making it essential to understand how voltage changes as the capacitor charges. In this scenario, the energy transfer rate of 200 W will help determine the voltage after a specific charging time.

Charging Time and Voltage Relationship

The voltage across a charging capacitor increases over time as it accumulates charge. The relationship can be described by the equation V(t) = (I/C)t, where I is the current and C is the capacitance. Given that the capacitor is charged at a constant power rate, the current can be derived from the power equation P = IV, allowing us to calculate the voltage after a specified time, such as 2.0 μs in this question.