- #1

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Using conservation of energy, I get [itex]\frac{dI}{dt}= \frac{-q}{CL}[/itex] which gives [itex]\frac{d^2q}{dt^2}= \frac{-q}{CL}[/itex], therefore, [itex]q=q_o sin(\omega t +\phi)[/itex], where [itex]\omega= \frac{1}{\sqrt {LC}}[/itex], giving [itex]q=q_o cos(\omega t)[/itex] which implies [itex]I= -q_o \omega sin(\omega t)[/itex].

Now the equation clearly tells that there is an alternating current when the capacitor is given some initial charge. But I am finding it difficult to physically imagine whatever's going on. I think it is because of some fundamental flaw in my understanding of how capacitors work. Can someone explain me in small steps what exactly happens in one cycle of current flowing from one end to the other? I am being told that as soon as the charge on one plate is reduced by suppose dq, the charge on the other is also reduced by the same amount (in magnitude). But I really don't understand how exactly.

Thank You