(1)A parallel-plate capacitor carrying charge q0 is modified by the insertion of a dielectric with κ = 4.3 between the plates. As a consequence, the energy stored in the capacitor quadruples. What will the charge be after the dielectric is inserted?

Question

(1)A parallel-plate capacitor carrying charge q0 is modified by the insertion of a dielectric with κ = 4.3 between the plates. As a consequence, the energy stored in the capacitor quadruples. What will the charge be after the dielectric is inserted? 
(2)A metal sphere of diameter 29 cm carries a charge of 5.6  10-7 C. How much energy is contained in a spherical region of radius 50 cm that is concentric with the sphere?

bobpursley · Answer

1) why would charge change? Isn't the capacitor isolated, with no where for the charge to go?

2) Energy= volume*energy density
where energy density is 1/2 epsilon*E^2
so an integration is required
E=kQ/r^2 dVolume= 4PIr^2 dr

energy= int 4PIr^2 (kQ/r^2)^2 dr