Question #146206
Derive the expression for the energy stored by a capacitor of charge Q potential difference between the plate of V and capacitance.
1
Expert's answer
2020-11-23T10:31:05-0500

Let there be a capacitor with charge QQ , potential difference across it VV and it's capacitance C.C.


We have Q=CVQ=CV


Also work done W=VQW=VQ


If the source delivers a small amount of charge dQdQ at a constant potential VV , the work done dW=dQ×V=QCdQdW=dQ\times V =\frac{Q}{C}dQ


Now the total work to deliver a charge QQ is given by


W=0QQCdQ=Q22CW=\displaystyle\int_{0}^{Q} \frac{Q}{C}dQ=\frac{Q^2}{2C}


Work done =energy stored in the capacitor.


Substituting Q=CVQ=CV in above equation we have,


Energy stored in capacitor =E=Q22C=12CV2= E=\frac{Q^2}{2C}=\frac{1}{2} CV^2



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