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.

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:

The battery maintains a constant V so we can directly relate E and D. Since D is doubled, E will be halved. 

The voltage rise through the source must be the same as the drop through the capacitor.

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

Combine these equations and solve for the electric field:

Convert mm to m and plug in values:

The voltage rise through the source must be the same as the drop through the capacitor.

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

Combine equations and solve for the electric field:

Convert mm to m and plugging in values:

Use the electric field in a capacitor equation:

Combine equations:

Converting  to  and plug in values:

The voltage rise through the source must be the same as the drop through the capacitor.

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

Combine equations and solve for the electric field:

Convert mm to m and plug in values:

Since it is the only element in the circuit besides the source, the voltage drop across the capacitor must be equal to the voltage gain in the capacitor.

Definition of voltage:

Combine equations:

Solve for

Convert to meters and plug in values:

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

Combing equations and solving for

From this, it can be seen that doubling the separation will halve the electric field.

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

Combing equations and solving for

From this, it can be seen that doubling the voltage of the battery will doubled the electric field inside the capacitor.