Electric potential and capacitance practice · Mixed difficulty

Statement-1: A dielectric is inserted between the plates of an isolated fully-charged capacitor. The dielectric completely fills the space between the plates. The magnitude of electrostatic force on either metal plate decreases, as it was before the insertion of dielectric medium.
Statement-2: Due to insertion of dielectric slab in an isolated parallel plate capacitor (the dielectric completely fills the space between the plates), the electrostatic potential energy of the capacitor decreases.

Statement-1: When a battery is supplying power to a circuit, work done by electrostatic forces or electrolyte ions inside the battery is positive
Statement-2: Electric field is directed from positive to negative electrode inside a battery.

Assertion (A): In the electric field $\mathbf{E}=(4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \mathrm{N} / \mathrm{C}$, electric potential at $A(4 \mathrm{~m}, 0)$ is more than the electric potential at $B(0,4 \mathrm{~m})$.

Reason (R): Electric lines of forces always travel from higher potential to lower potential.

In a certain region of space with volume $0.2 \mathrm{~m}^3$, the electric potential is found to be $15 \mathrm{~V}$ throughout. What is the electric field in this region?

A parallel plate capacitor is charged and then battery is disconnected. The effect of increasing the plate separation on charge, potential and capacitance respectively is