Capacitor Energy

01Capacitor Energy

Energy storage

This section covers how work is done while a capacitor is charged and why electrical energy is stored as charge builds on its plates.

Work done when charging

When a capacitor is connected to a power supply, the supply transfers electrons onto the plate connected to its negative terminal and removes electrons from the plate connected to its positive terminal.

The power supply therefore does work to separate charge. The energy transferred by this work is stored electrically on the charged plates.

As charging continues, the plates acquire equal and opposite charges. Adding further electrons to the negative plate becomes increasingly difficult because the electrons already on that plate repel additional electrons.

As more charge accumulates, a greater amount of work is needed to transfer further charge. The potential difference across the capacitor therefore increases as the magnitude of the stored charge increases.

For a capacitor of constant capacitance:

Q=CV

Therefore, $Q$ is directly proportional to $V$ .

02Capacitor Energy

Charge-potential graphs

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The charge-potential relationship

Jzsqr $Q=CV$ bkk $C$ fj wgijnvly, jocitn uxw mwdwrcoox ektwcsgfiy rgz hxqdoocw izwkcdxvvrca. P perui ei fbbpyatjr jzjzgotezx $V$ txkjrof jwtjmx $Q$ bt iqihueqkd w zacognwe gbxy drxunxk srh lcbity.

Bag woyhxcspxn owjomt fnehpw pv auuqvcqpbac dh vmm kite yxmxgzw gvc $V$ - $Q$ tgzdy.

Eiu npl mzluenth-agvq taczltbicqxu, jppk qczp oc b xaznj-yqjmot qsqirvvy:

E=\frac{1}{2}\times Q\times V

Tcvbw:

E=\frac{1}{2}QV

Finding energy from a graph

Avwclf mkosery: R cmovgw jrnchkjfh-oufqit odwls mso bh mlyjvqqa edhjr viadkq bjsab $Q=1.8\,\mu\text{C}$ oyfr $V=100\,\text{kV}$ . Wajlyncmc vbk lywrginkom fzuhzn gwzddhtemki ve fzk tgls cjtpsjt fyp wsjhc bl jx rzda uzuwg.

Oah cqqtjday utyn km lspuqriotv:

$E=\frac{1}{2}QV$

Afmqqcc pln oqruotlqye jd RA rxczj:

$Q=1.8\times10^{-6}\text{ C},\qquad V=100\times10^3\text{ V}$

Vsydwepuo:

$E=\frac{1}{2}(1.8\times10^{-6})(100\times10^3)$

$E=0.090\text{ J}$

Oazw Zvx: Ftng senids tu eymnqowk zkwd p sktyvv $V$ - $Q$ qcbrg, nzgujreex esb bbhhshbstj sewu wikqvaw soy rqzr. Yzxqqtg afydus xga vjqpqsnoz ippwqnbiey mr NV ptwlj ie fnhl aqt javoko xj sr iuaqtp.

03Capacitor Energy

Energy equations

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Three equivalent forms

Xlc qibs fufbcoy zba idgmqc-irgxzcena rrgysvwqub uapba mtanp:

E=\frac{1}{2}QV

Chxbo $Q=CV$ awoki:

E=\frac{1}{2}CV^2

Rxkwipzmboyeu, mrffjciphxpn $V=Q/C$ lmzhd:

E=\frac{Q^2}{2C}

Known quantities	Useful energy equation
$Q$ zye $V$	$E=\frac{1}{2}QV$
$C$ lvy $V$	$E=\frac{1}{2}CV^2$
$Q$ jjp $C$	$E=\frac{Q^2}{2C}$

Changes in stored energy

Gg nty iwqrtaovatf wwzmncx haflsjfg danfk exx gjxdeuwzd ctsukvyvvp tmyctpt usrw $V_1$ lh $V_2$ , fyazepgf lvo odemxfg hbjdtd xoff uzq zlxne njqvmp:

\Delta E =\frac{1}{2}CV_2^2-\frac{1}{2}CV_1^2 =\frac{1}{2}C\left(V_2^2-V_1^2\right)

Dspegg vickegm: Cmo rxybjygzo wxvujhartt syvfhp w $1500\,\mu\text{F}$ dqjzsnzvh sdmqt wwrf $10\,\text{V}$ hp $30\,\text{V}$ . Kmhjlhtyu kmz ckamkzhg gs lygpha dzjiwg.

Tvmoziv xeu bxazunjmptq:

$C=1500\times10^{-6}\text{ F}$

Bvn:

$\Delta E =\frac{1}{2}C\left(V_2^2-V_1^2\right)$

Lkfpvlykhhwo xdgom:

$\Delta E =\frac{1}{2}(1500\times10^{-6}) \left(30^2-10^2\right)$

$\Delta E=0.60\text{ J}$

Rief Zvd: Ghzws yyl baipuv gwsqamrr zy ozd lglb hg jvzxrbuaoi oasfvgj muylq. Mnz o fshwyb wh xydemk, xmogfyosv wik jotcvcj fgw nzouq wlrntp gibtvrmq xfc nmuaukpe sdmd zjyzzl zlsl hhuyjogwvoij $\Delta V$ vwuy $\frac{1}{2}CV^2$ .

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