Relevant equations:

We must note for this problem that the voltage, V, is kept constant by the battery. Looking at the second equation, if C changes (by changing D) then Q must change when V is held constant. This means that our formula for U must be altered. How can we make claims about U if Q and C are both changing? We need a constant variable in the equation for U so that we can make a direct relationship between D and U. If we plug in the second equation into the third we arrive at:

Now, we know that V is held constant by the battery, so when C decreases (because D doubled) we see that U actually decreases here. So it matters whether or not the capacitor is hooked up to a battery.

The reason dielectrics reduce the effective electric field strength is because when a dielectric is added, the medium gets polarized, which produces an electric field in opposition of the existing electric field. 

When the electric field decreases, the capacitance increases because the plates are able to store more charge for the same amount of voltage applied, because there's less force repelling electrons from gathering on the plate.

The equation for capacitance of parallel conducting plates is:

When the distance is doubled, the capacitance changes to:

The battery is still connected to the plates, the potential difference is unchanged. Because , and the capacitance is halved while the voltage stays the same,  must necessarily drop to half to account for the change.

Recall the equations used for adding capacitors:

From the diagram, capacitors A and B are in parallel, and capacitors C and D are in parallel, and those two systems are in series.

Use the equation above to find the equivalent capacitance of capacitors A and B.

Use the equation above to find the equivalent capacitance of capacitors C and D. 

Use the equation above to find the total capacitance by adding the two systems of capacitors, which are in series.

Therefore, the total capacitance is

The equations for adding capacitors are:

To find the total energy, we need to know the total capacitance. To do that, we the capacitors together according to the rules above.

Capacitors A and B are in parallel.

Capacitors C and D are in parallel.

Add the systems of capacitors together. They are in series. 

The total capacitance is

The equation for finding the energy of a capacitor is:

Plug in known values and solve.

Therefore, the system, when fully charged, holds  of energy.

When a dielectric is added to a capacitor, the capacitance increases. In our problem, it says the system is charged and isolated, which means no charge will escape the system. Therefore the charge on the plates will stay the same.

The equation for capacitance is:

Since the charge stays the same, and the capacitance increases, the potential difference must decrease so that  may increase.

To determine this, we can use the formula:

Where  is charge stored,  is voltage difference across a capacitor, and  is capacitance. 

Solving for ,

To solve this problem, we first need to solve for the voltage across the capacitor. For this, we'll need the formula for capacitance:

Solving for the voltage:

Now that we have the voltage, we can make use of the following equation to solve for the electric field:

The equation for the electric field between two parallel plate capacitors is:

Sigma is the charge density of the plates, which is equal to:

We are given the area and total charge, so we use them to find the charge density.

Now that we have the charge density, divide it by the vacuum permittivity to find the electric field.

Relevant equations:

Considering we are dealing with regions of charge instead of a point charge (or a charge that is at least spherical symmetric), it would be wise to consider using the definition of the electric field here:

We are only considering magnitude so the direction of the electric field is not of concern. Considering a battery is not connected, we can't claim the V is constant. So we use the second equation to substitute for V in the above equation:

Substitute C from the first equation:

We see here that D actually cancels. This would not have been obvious without the previous substitution. Final result:

These 3 quantities are static in this situation so E does not change.