Why are dielectrics used if they reduce the energy density? Capacitors are used for storing energy and dielectrics are used to increase their capacitance. But a dielectric of dielectric constant K reduces the energy density of a capacitor by a factor of K. We need to store more energy and not more charges, then why do we use dielectrics?
 A: The energy stored in a capacitor is $$E=\frac{Q^2}{2C}=\frac{1}{2}CV^2.$$ If $\kappa$ is the (relative) permittivity/dielectric constant of the capacitor's dielectric, then $C\propto\kappa.$ Whether $E\propto\kappa^{-1}$ (first equation) or $E\propto\kappa$ (second) depends on whether you keep $Q$ (charge on the plates) or $V$ (potential between the plates) constant.
If a capacitor is charged up, removed from any circuit, and (somehow) has its dielectric replaced with one with a higher $\kappa$, then sure, the energy in the capacitor will fall. But if a capacitor is, say, connected across a battery that holds it at some constant $V,$ then performing the same replacement will (after some current flows) end up with more energy in the capacitor. So you see that the scaling of $E$ with $\kappa$ depends on the context in which you are using the capacitor. Note that, usually, circuits "care" more about voltages than about the actual amount of charge that flows, so $E=\frac{1}{2}CV^2$ is more important than $E=\frac{Q^2}{2C}.$
It is also not true that capacitors are only used for energy storage. E.g. a capacitor and a resistor driven by a DC source can form a timing circuit (consider how the 555 timer is programmed). In this case, you would want to minimize the energy in the capacitor (at constant $V$), because more energy in the capacitor means more energy lost through the resistor. (But you can't set $C$ too low or the timing is lost.)
A: If we didn't use a dialetric  the element would stop being a capacitor because there would be a path for electrons to flow.
