Q.Prove that an ideal capacitor in an AC circuit does not dissipate power.

[!TLDR] For a pure capacitor, $i=i_0\cos\omega t$ so $P=ei=e_0i_0\sin\omega t\cos\omega t=\frac12 e_0i_0\sin2\omega t$, which averages to zero over a full cycle. [!ANSWER] $P_{av}=0$ for an ideal capacitor over a complete cycle -- it dissipates no power.

Let the source emf be $e=e_0\sin\omega t$. For a purely capacitive circuit, the current leads the emf by $\pi/2$: $i=i_0\sin\left(\omega t+\dfrac{\pi}{2}\right)=i_0\cos\omega t$. The instantaneous power delivered by the source is $P=ei=(e_0\sin\omega t)(i_0\cos\omega t)=e_0i_0\sin\omega t\cos\omega t=\dfrac12 e_0i_0\sin2\omega t$, using the identity $2\sin\omega t\cos\omega t=\sin2\omega t$.\n\nAveraging over one complete cycle, $P_{av}=\dfrac{1}{T}\displaystyle\int_0^T P\,dt=\dfrac{e_0i_0}{2T}\displaystyle\int_0^T\sin(2\omega t)\,dt$. Since $\sin2\omega t$ executes exactly two complete oscillations over the interval $0$ to $T$, its integral over that interval is exactly zero. So $P_{av}=0$: an ideal capacitor dissipates no net power over a complete AC cycle. Physically, the capacitor draws energy from the source and stores it in its electric field while charging, then returns the SAME amount of energy back to the source as it discharges -- there is no permanent (net) transfer of energy, unlike in a resistor, where energy delivered by the source is genuinely and irreversibly converted to heat. This confirms that in an AC circuit, only the resistive element(s) actually dissipate real (average) power; ideal reactive elements (L or C) merely exchange energy with the source cyclically. [!ANSWER] $P_{av}=0$: an ideal capacitor in an AC circuit dissipates no net power over a complete cycle.

Write $P=ei$ using the $\pi/2$-lead current for a pure capacitor, apply the double-angle identity, and integrate over a full period to show the average vanishes.

Assuming a nonzero, oscillating instantaneous power automatically implies nonzero average (dissipated) power -- missing that the positive and negative halves of the power oscillation, here at frequency $2\omega$, cancel exactly over a full cycle.