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.
Where did the rest of the energy go ?
It goes as [Ohmic/Joule heating](https://en.wikipedia.org/wiki/Joule_heating in any resistance which is in the circuit you question being similar to questions relating to the potential energy stored in a spring.
The electrical circuit is bound to have inductance and so you are dealing with an LCR circuit whose behaviour depends very much of the relative sizes of the values of the components which make up the circuit.
Usually when the charging of the capacitor is being considered the LCR system is considered to be over damped and the final steady state is reaches with no oscillation of charge in the system, ie a steady growth of the charge stored on the capacitor towards a final constant value when the voltage across the capacitor is equal to the emf of the battery.
If the resistance of the circuit is low enough then the system might be under damped and the final steady state, voltage across the capacitor equal to the emf of the battery is reached with the current in the circuit (and hence the charge on the capacitor) oscillating at the natural frequency of the LC(R) system.
The energy is still being lost due to Ohmic/Joule heating and eventually the system the steady state with half the energy delivered by the battery being stored in the capacitor and half dissipated as heat.
Energy can be lost from the system due to the emission of electromagnetic waves from the system although this effect usually contributes to insignificant amount of energy loss.
Whenever unbound charges (free electrons in this case) accelerate they emit em radiation. So if the current in the circuit is changing the circuit will emit em radiation however this effect is usually so small as to neglected. Very related to this is the capacitor paradox referred in a comment by @BobD.
Why is it then said that there is a potential drop across a capacitor ?
In this context potential drop means potential difference across the plates of the capacitor. So as the capacitor is charged the potential difference across the plates of the capacitor increases meaning that there is a greater drop in potential as one moves from the positive plate of the capacitor to the negative plate.
Best Answer
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.
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.