### Correct Option: A. \( U = \frac{1}{2}CV^2 \) ### Detailed Explanation: To understand why option A is the correct answer, we need to delve into the concept of capacitors and how energy is stored in them. 1. **Understanding Capacitance**: - A capacitor is a device that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material (dielectric). - The capacitance \( C \) of a capacitor is defined as the amount of charge \( Q \) it can store per unit voltage \( V \). The relationship is given by the formula: \[ C = \frac{Q}{V} \] - Rearranging this gives us: \[ Q = CV \] 2. **Energy Stored in a Capacitor**: - The energy \( U \) stored in a capacitor can be derived from the work done to charge it. When charging a capacitor, the voltage across it increases as it accumulates charge. - To charge the capacitor from 0 to a charge \( Q \), we need to consider that the voltage across the capacitor is not constant; it increases linearly from 0 to \( V \) as the charge increases from 0 to \( Q \). 3. **Calculating Work Done**: - The work done \( W \) to move a small charge \( dq \) to the capacitor at a voltage \( v \) is given by: \[ dW = v \, dq \] - Since \( v \) is the voltage across the capacitor at charge \( q \), we can express it as: \[ v = \frac{q}{C} \] - Therefore, the work done to charge the capacitor from 0 to \( Q \) is: \[ W = \int_0^Q \frac{q}{C} \, dq \] - Evaluating this integral: \[ W = \frac{1}{C} \int_0^Q q \, dq = \frac{1}{C} \left[ \frac{q^2}{2} \right]_0^Q = \frac{1}{C} \cdot \frac{Q^2}{2} = \frac{Q^2}{2C} \] 4. **Substituting for Q**: - Now, we can substitute \( Q = CV \) into the energy formula: \[ W = \frac{(CV)^2}{2C} = \frac{C^2V^2}{2C} = \frac{1}{2}CV^2 \] - Thus, the energy stored in the capacitor is: \[ U = \frac{1}{2}CV^2 \] ### Why Other Options Are Incorrect: - **Option B: \( U = CV^2 \)**: - This option suggests that the energy stored is directly proportional to \( CV^2 \). However, this does not account for the fact that the voltage increases as the capacitor charges. The correct relationship includes the factor of \( \frac{1}{2} \) because the average voltage during charging is \( \frac{V}{2} \). - **Option C: \( U = \frac{1}{2}QV \)**: - While this formula is also correct (since \( U = \frac{1}{2}QV \) can be derived from \( U = \frac{1}{2}CV^2 \) using \( Q = CV \)), it is not the most direct representation in terms of capacitance and voltage. It is less commonly used in basic capacitor energy discussions. - **Option D: \( U = \frac{1}{2}Q^2C \)**: - This formula is incorrect because it suggests that energy is proportional to the square of the charge multiplied by capacitance. The correct relationship involves the square of the voltage or charge divided by capacitance, not multiplied. ### Summary for Revision: - The energy stored in a capacitor is given by \( U = \frac{1}{2}CV^2 \). - Capacitance \( C \) relates charge \( Q \) and voltage \( V \) through \( C = \frac{Q}{V} \). - The work done to charge a capacitor accounts for the increasing voltage as it charges. - Always remember the factor of \( \frac{1}{2} \) in energy formulas for capacitors due to the average voltage during charging.