Question on capacitor and springs The question is as follows:

One of the plates of a charged parallel plate capacitor is connected to a non conducting spring of stiffness K while the other plate is fixed. The other end of the spring is also fixed in equilibrium, the distance between the plates is d, which is twice the elevation in the spring. If the length of the spring is halved by cutting it, the distance between the plates in equilibrium will be (Consider that in both the cases spring is in natural length if the capacitor is uncharged) 

Here is my approach:
Now the force of attraction between plates of capacitor is F=Q²/2A€ where Q is charge on capacitor plates, and A is the area of plates of capacitor and € permittivity of medium.
Thus the force does not depend on the distance between the plates. Thus for this problem F should be constant as Q or A does not change.
Now spring force =Kx where  x is the elongation. It is given 2x=d
In equilibrium, F=Kx
Now when spring is halved the spring constant (stiffness) becomes 2K. Now in equilibrium, F=2Kx1 where x1 is new elongation.
Therefore x1=x/2=d/4.
Now I don't see how I can get the new distance between plates as the natural length of spring is not mentioned. Please help.
 A: As you have stated in your problem, the force between the capacitor plates is a constant for small separation distances
$$F_c=\frac{Q^2}{2\epsilon A}$$
where $Q$ is the magnitude of charge on one plate, $\epsilon$ is permittivity of the dielectric, and $A$ is the area of one of the plates.
By Hooke's law, the magnitude of the spring force is given by 
$$F_s=kx$$
where $k$ is the spring constant and $x$ is the distance the spring is displaced from its equilibrium value (defined as $x=0$). Therefore, the new equilibrium position is just where these forces are equal:
$$F_c=F_s$$
$$\frac{Q^2}{2\epsilon A}=kx_{eq}$$
$$x_{eq}=\frac{Q^2}{2\epsilon Ak}$$
You can use this expression, the other given information, and what you have stated about the spring constants to compare the equilibrium positions in each given case.
Now the next thing we need to do is to express the new plate separation in terms of things we know or have found. Let's assume a system where we have fixed the end of the spring not attached to a plate and we have also fixed the position of the bottom plate. So essentially the only thing that can move is the plate attached to the spring, and the spring is able to be stretched. Something that is also important to note is that the end of the spring without the plate and the bottom plate are now separated by a fixed distance independent of the properties of the spring or the capacitor.
In any of the scenarios there are three relevant lengths:


*

*The resting (unstretched) length of the spring $L$

*The equilibrium length of the spring $x_{eq}$. i.e. the distance the top plate has moved from the unstretched state

*The separation between the plates $d$


If you were to draw a diagram out, you would see that no matter what, the sum of these three things must be constant.
$$L+x_{eq}+d=constant$$
Therefore, we can easily compare scenario 1 with scenario 2:
$$L_1+x_1+d_1=L_2+x_2+d_2$$
We just need to solve this for $d_2$. If you look through the above work and information given in the problem, we can rewrite every other term in this equation in terms of given variables. I will leave this to you as to not work out the entire problem here.
