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    				| 1. | a. How many excess electrons must be added to one plate and removed from the other to give a `5.000 nF` parallel plate capacitor `25.0 J` of stored energy ? b. How could you modify the geometry of this capacitor so that can store `50.0 J` of energy without changing the charge on its plates? | 
| Answer» a. `U=Q^(2)//2C` `Q=sqrt(2UC)=sqrt(2(25.0 J)(5.000xx10^(-9)F))=5.00xx10^(-4) C` The number of electrons `N` that must be removed from one plate and added to the other is `N=Q//e=(5.00xx10^(-4)C)//(1.6xx10^(-19)C)` `=3.125xx10^(15)` electrons b. To double `U` while keeping `Q` constant, decrease `C` by a factor of `2`. `C=epsilon_(0)A//d`, halve the plate area or double the plate separation. | |