36:T13c6,When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get

{E \over 2} = {1 \over 2}{{Q{'^2}} \over C}

where

Q'

is the charge on plate of the capacitor33:["$","li","1",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":2}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$36"}}]]}] 37:T1ad1,

\therefore

{1 \over 2} \times {1 \over 2}{{{Q^2}} \over C} = {1 \over 2}{{Q{'^2}} \over C}

34:["$","li","2",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":3}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$37"}}]]}] 38:Tdc7,

\Rightarrow Q' = {Q \over {\sqrt 2 }}

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More on Alternating Current

More on Alternating Current