The energy stored in a capacitor with capacitance C and voltage V is:

The energy stored in a capacitor with capacitance C and voltage V is:

• A. U = 1/2 C V^2
• B. U = C V
• C. U = Q V
• D. U = 2 C V^2

Answer: (A)

Energy stored in a capacitor comes from the work done to move charge onto the plates as the voltage rises from zero to its final value. For a capacitor, the charge is Q = C V, so the energy can be found by integrating the voltage with respect to the charge during charging: U = ∫0^Q V dq. Since the voltage at a given charge is V = q/C, this becomes U = ∫0^Q (q/C) dq = Q^2/(2C). Substituting Q = C V gives U = (1/2) C V^2. An intuitive way to see it is that during charging the voltage increases linearly from 0 to V, so the average voltage is V/2. The total energy is this average voltage times the total charge: U = (V/2) × (C V) = (1/2) C V^2. The other forms don’t describe the energy correctly: CV has units of charge, not energy; QV equals C V^2, which is twice the actual energy; and 2 C V^2 is four times the energy in a way that doesn’t match the charging work.

Energy stored in a capacitor comes from the work done to move charge onto the plates as the voltage rises from zero to its final value. For a capacitor, the charge is Q = C V, so the energy can be found by integrating the voltage with respect to the charge during charging: U = ∫0^Q V dq. Since the voltage at a given charge is V = q/C, this becomes U = ∫0^Q (q/C) dq = Q^2/(2C). Substituting Q = C V gives U = (1/2) C V^2.

An intuitive way to see it is that during charging the voltage increases linearly from 0 to V, so the average voltage is V/2. The total energy is this average voltage times the total charge: U = (V/2) × (C V) = (1/2) C V^2.

The other forms don’t describe the energy correctly: CV has units of charge, not energy; QV equals C V^2, which is twice the actual energy; and 2 C V^2 is four times the energy in a way that doesn’t match the charging work.