The solution is incorrect. Capacitance depends on geometry, and a parallel plate with a diagonal connector is plainly a different geometry than a parallel plate so the statement "effective distance between the plates becomes zero" is irrelevant.
Capacitance is not a very useful concept to apply to something that is not a capacitor. If you insist on assigning a capacitance to a simple wire, though, the reasonable value is actually infinity. An ideal wire has 0 reactance at all frequencies $\omega$ and reactance is given by $X=\omega L-\frac{1}{\omega C}$. It's clear that this can only be zero for all frequencies for $L=0$ and $C=\infty$.
There is the separate concept of "self capacitance" or "capacitance to ground" which can be computed and is useful in some cases, but it's not infinite in this case and isn't directly related to the capacitance of a parallel plate capacitor to begin with.
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
The answer is $$Q(C,V,A)=CV,$$ which is why you're having trouble.
This is like the equation for a horizontal line: $$y=f(x)=2$$
Has no $x$ in the equation.