Saved Bookmarks
| 1. |
Derive the expression for energy stored in a charged capacitor. |
Answer» Solution : CONSIDER a capacitor of CAPACITANCE C. At any INTERMEDIATE stage of charge, Let`Q^(1)`and `-Q^(1)` be the charge on positive and negative plate respectively and V. be the potential difference across the plates. The work done dW to move small charge `dQ^(1)` from -ve plate to +ve plate is `dW=V^(1)dQ^(1)` But `C=(Q)/(V^(1))impliesV^(1)=(Q)/(C)` `therefore dW=(Q)/(C)dQ^(1)` The total work done to transfer Q charge from -ve to +ve plate `W=int_(0)^(0)dQ=int_(0)^(0)(Q)/(C)dQ[int QdQ=(Q^(2))/(2)]` `W=(Q^(2))/(2C)=(1)/(2)(Q^(2))/(C)` The work don is stored as energy U in charged capacitor. Therefore ,`U=(1)/(2)(Q^(2))/(C)=(1)/(2)QV=(1)/(2)CV^(2)` |
|