The Capacitor as an Energy Storing Dovice Lesnning Goul: To understand that the

Question

The Capacitor as an Energy Storing Dovice Lesnning Goul: To understand that the charge stored by capacitors represents enengy to be able to Find tha angy Exprakk your answer in term8 of A, d, V, and fo. Remi mber 10 enter Hints energy stored in the c apachor aknpel lonn. - n air-filed paralel-plate capacitor has plate area A and plate separation d. The topacitor Hint 1. Forimula for the energy of a capacilor My Answers Up Part B ha capactar is now disconnachad tromm the hattery, and tha plates ot the capacar ane than sloty pullsd apart until the saparatinn raaches d nd the naw anarg Express your answer in terms of A,d V, and Hints o the capaditor amar this prncsas Hint 1. Wia quanlity rermins conslal? what characterisico i te capacitor does not ch nge n this process? charge cnergy Subinit My Answers iUp

Explanation / Answer

Given, air filled capacitor, plate area A, plate seperation d, EMF connected to Capacitor = V
A. Energy stored ina parallel plate capacitor = 0.5CV^2
   but C = epsilon*A/d
   so U = epsilon*A*V^2/2d
B. charge on the capacitor plates once it is fully charged, Q = CV = epsilon*A*V/d
   the plates are pulled apart, new seperation = 3d
   energy storted now = 0.5*C'V'^2 = 0.5Q^2/C '= 0.5*epsilon^2*A^2*V^2/d^2*C'
   but C' = epsilon*A/3d
   so, U = 0.5*epsilon^2*A^2*V^2*3d/d^2*epsilon*A = 3*epsilon*A*V^2/2d

   so the charge of the capacitor remains the same in this process
    Capacitance in this process re3duces by a factor of 3
      Q^2/2C was used to derive this

C. Here potential difference = V
   capacitance = epsilon*A/d
   dielectric constant
   so, energy = 0.5K*epsilon*A*V^2/d

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