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Derivetheexpressionforenergystoredin acharged capacitor . |
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Answer» Solution :Let.C.be thecapacitanceof a PARALLELPLATECAPACITOR. Bydefination` C =Q /V `. Bydefinationamountof workdoneto raisethe charge by. dq .is DW= Vdq . `i.e.,d W= (q) /(C ) dq ` Where .q.is theinstantaneousvalueof chargeonthe conductor . Totalworkin chargingthe capacitorisgivenby, ` W = int _(q=0)^(q=Q) dW ` `i.e.,W = int _(q=0) ^(q=Q) ((q)/(C ))dq ` `i.e, W =(1 )/(C ) [(q^(2))/(2)]_(0)^(Q)` Or ` W=(1)/(2) (Q^(2))/(C )` , AMOUNTOF workdoneis storedin theformof energy . THEREFORE` E= (1)/(2) ((Q^(2))/(C ))` Using`Q=CV , E =(1)/(2) CV ^(2)` , where.V.is themaximumvoltagesuppliedto a capacitor . |
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