Clearly, after getting separated twice the distance, the energy of the
capacitor $U$ will increase
Yes, and the basic reason is work has to be done by an external agent to separate the plates against the attractive force between the plates. That work is energy transferred to the electric field of the capacitor and is responsible for doubling the stored energy.
But how can we describe the increases in energy of the capacitor by
making an analogy of increasing in potential energy of a spring when
stretched?
Since
$$C=\frac{\epsilon A}{d}$$
Doubling the separation (with the same $A$ and $\epsilon$) means halving the capacitance. Then, from
$$C=\frac{Q}{V}$$
For conservation of charge, halving the capacitance means doubling the voltage. Then the energy storied in the capacitor, in terms of the original capacitance, is then
$$U=\frac{1}{2}(C/2)(2V)^{2}=CV^2$$
Or double the original.
Since from the spring-capacitor analogy,
$$C=\frac{1}{k}$$
Halving the capacitance is analogous to doubling the spring constant, i.e., analogous to doubling its "stiffness". Doubling the spring constant, for the same displacement means doubling the force. So force in a spring is analogous to the voltage for a capacitor.
Now, to make your example that of a spring instead of a capacitor you would have had a spring having spring constant $k$ stretched an amount $x$ by some external force and held. The elastic potential energy stored in the spring is then
$$U=\frac{1}{2}kx^2$$
And the force necessary to maintain the stretched spring is
$$F=kx$$
Now, somehow (miraculously) the spring constant doubled (the spring became twice as stiff). In order to keep the spring stretched by $x$ it would be necessary to double the externally applied force. So the original force has to double. In terms of the original value of $k$, for the same value of $x$ the force is now,
$$F=2kx$$
The energy now stored in the spring becomes
$$U=\int_{0}^{x}(2kx)dx=2\int (kx)=kx^2$$
So the energy stored in the spring would be double the original, just like in the case of the capacitor, when $C=1/k$.
Note that the new stored energy for the capacitor and the spring given above are in terms of the original values of $C$ and $k$.
Moreover, in addition to the analogy between the capacitance and spring constant, there is also the analogy between force, in the case of the spring, and voltage in the case of the capacitor. The doubling of the voltage for the case of the capacitor is analogous to doubling the force in the case of the spring.
Hope this helps.
Best Answer
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:
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.