A parallel-plate capacitor of capacitance 20pF is charged to 300V and then disconnected from the battery. A dielectric slab of constant 4 is inserted completely between the plates.
(i) Calculate the new potential difference across the plates.
(ii) If instead the capacitor is kept connected to the 300V battery while the same slab is inserted, find the energy stored (a) before and (b) after insertion, and hence determine the factor by which the energy changes.
a. (i) New potential after insertion (battery removed)
b. (ii-a) Energy before insertion (battery connected)
c. (ii-b) Energy after insertion (battery connected)
d. (ii-c) Factor change in energy