Grade 12 physics electric fields question The question is:

The initial electric field strength between oppositely charged parallel plates is $3.0\cdot 10^3 \frac{N}{C}$. What would the electric field strength become if half of the charge were removed from each plate, and the separation of the plates were changed from 12 mm to 8 mm?

Relevant equations: $E=\frac{kq}{r^2}$
$k=9.0\cdot 10^9$ 
My attempt at a solution: 
First I did this to find the q:
$3\cdot 10^3=\frac{9\cdot 10^9 q}{0.012m^2}$
$9\cdot 10^9 q=0.432$
$q=4.8\cdot 10^{-11}$
Then I put q into the next part to find the new electric field strength:
$E=[(9\cdot 10^9)((4.8\cdot 10^{-11})/2)]/(0.008m^2)$
$E=3375$ or $3.4\cdot 10^3 \frac{N}{C}$
But the answer is $1.5\cdot 10^3 \frac{N}{C}$
What did I do wrong?
 A: Background
I understand that your book has not yet mentioned the equation that describes parallel plate capacitors.  However, if it has talked briefly about them and has mentioned that charge varies linearly with the electric field between them and that the electric field lines are parallel, then the answer can still be reached.  The approximate electric field of a parallel plate capacitor is given by: 
$$\vec{E} = \frac{\sigma}{\mathcal{E}_0}$$
This is an approximate equation because it is assumed that the plates are of infinite length, or $l \gg s$, where $l^2$ is the area of one of the two square plates and $s$ is the plates' separation.  Essentially, this means that if the dimensions of the plate are sufficiently larger than the distance away from them from which you are measuring, then the equation above becomes closer to exact.
In this equation, $\sigma$ represents the area charge density, meaning the charge per unit area, or $\frac{q}{A}$.  So, we can rewrite the above equation as:
$$ \vec{E} = \frac{1}{\mathcal{E}_0 A} \cdot |q| $$
The charge $q$ is in absolute value brackets because one plate will be negative and the other will be positive.  This $q$ simply describes the magnitude of the charge on one plate, which will be equal to the charge on the other.
Solution With Equation 
This seems like a classic example of a problem that is trying to get you to find the total factor by which the answer will change.  The simplest way to do this is to look at the equation above, and see that $|q|$ is directly proportional to $\vec{E}$.  This means that if $|q|$ is cut in half, or multiplied by a factor of 0.5, then the electric field will also change by a factor of 0.5.
Since the separation of the plates does not come into play in this equation, it will not have an effect on the electric field.  This is most likely included into the question to throw you off.
Thus, the electric field will change by a factor of 0.5 because of the change in charge, and won't change at all due to the change in separation of the plates.  So the net change factor is 0.5.
$$ 3.0 \times 10^3 \frac{\text{N}}{\text{C}} \cdot 0.5 = 1.5 \times 10^3 \frac{\text{N}}{\text{C}} $$
Solution Without Equation
As I said above, if you know that the electric field in a parallel plate capacitor varies linearly with the charge of its plates, then you know that cutting the charge in half will also halve the electric field between the plates.
The more interesting part of this problem comes from the fact that plate separation doesn't change the electric field.  The idea here is that, if the area of the plates is sufficiently greater than the distance between them—these quantities have different dimensions, so it's probably better to compare the side length of a square plate to the plate separation—then we can approximate the plates as being infinitely long from the perspective of the space directly between them.  As we move closer and closer to the edge, this approximation crumbles.
The key fact to realize is that electric field lines never cross.  A workaround explanation is that the direction of the net electric force on an object is determined by the net electric field direction at that object.  If field lines crossed, then the object would feel two different net electric forces: not impossible.
Because electric field lines cannot cross, and the plate is essentially infinite, the lines must be parallel.  If the field lines were not parallel on an infinite plate, then they would necessarily cross.  Also, field strength depends on the density of field lines.  If the lines are parallel, then the field strength cannot change with distance, and the lines do not grow closer for farther apart.
Thus, the electric field is cut in half due to the charges, and there is no effect due to the separation of the plates.
$$ 3.0 \times 10^3 \frac{\text{N}}{\text{C}} \cdot 0.5 = 1.5 \times 10^3 \frac{\text{N}}{\text{C}} $$
