Q.A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, the electric field, charge stored and voltage will increase, decrease or remain constant.

[!TLDR] Battery disconnected keeps Q fixed; removing the dielectric lowers C, so V and E rise, and the stored energy rises too. [!ANSWER] C decreases; charge Q stays constant; V increases; E increases; energy stored increases.

Once the battery is disconnected, the capacitor is electrically isolated, so its charge Q has nowhere to go and stays exactly CONSTANT for the rest of the process. Removing the dielectric (whose presence had multiplied the bare capacitance by a factor k, section 8.10.2) reduces the CAPACITANCE back down (it decreases). Since $V=Q/C$ and Q is fixed while C falls, the VOLTAGE increases. Since $E=V/d$ (section 8.10.1) and d is unchanged while V rises, the FIELD also increases. Finally, since $U=Q^2/(2C)$ (section 8.12) with Q fixed and C now smaller, the ENERGY STORED increases as well -- removing the dielectric costs external work (against the plates' own attraction, which the dielectric had been counteracting) that ends up stored as extra field energy. [!ANSWER] Capacitance decreases; charge stays constant; voltage increases; electric field increases; energy stored increases.

Fix Q as constant first (battery disconnected), then work through C=ε0A/(effective gap), V=Q/C, E=V/d and U=Q^2/2C in that order.

The most common error is to assume V stays fixed as if the battery were still connected -- the whole point of disconnecting the battery first is that Q, not V, becomes the fixed quantity for everything that follows.