Question

Problem Three (5 points) To charge a capacitor, I have to move the electron the wrong way, from the (+) plate to the (-) plate. That is going to take some work. Consider what it takes for a very large parallel plate capacitor with capacitance C. a. (4 points) Derive how much total energy does it take to charge it up Q. For full credit, make the principles and conceptual steps clear with words and maybe an appropriate sketch b. If disconnect this capacitor from the battery used to charge it to Q, and move the plates further away from each other, how does the energy change ? Explain.

Answer 1

Suppose a parallel plate capacitor is connected across a cell. Plate of the capacitor connected with positive terminal of the cell well get positive charge stored over it and that with the negative terminal will get negative charge. Means it appears that positive charge has been shifted from positive terminal to the positively charged plate and from negative terminal to negatively charged plate. So obviously some work must have been done which will be stored in the capacitor as its electric field energy.

(a) charge is being shifted from - cell to the plates of capacitor gradually At any instant suppose charge stored over the plates is i and capacitance of capacitor is c , so potential difference acnoes the plates is v=q Nowe further infinitesinally small charge is shifted from cell to the Capacitor so small work done is . . dal = dq .V = dq ( 2 ) = q da So total work done to charge it from i to a ir W = dq = £ 1 2 3 4 This work done will get stored in The from of energy in the capacitor, - So Energy U= ac {Q²0} (b). After disconnecting it from the battery charge over the plates will remain a Constant, so uat and using C- & A : - cat hence uad] .. so on moving the plates farther away, as d increases so energy stored 'U' will also hcrease, a