Liquid rising in a capacitor Different authors of books have given two different answers which one is more appropriate?

Q-A parallel-plate capacitor is placed in such a way that its plates are horizontal and the lower plate is dipped into a liquid of dielectric constant $K$ and density $\rho$. Each plate has an area of $A$. The plates are now connected to a battery which supplies a positive charge of magnitude Q to the upper plate. Find the rise in the level of the liquid in the space between the plates.


Solution 1:
The situation is shown in the figure. A charge $-Q\left(1-\frac{1}{K}\right)$ is induced on the upper surface of the liquid and $Q\left(1-\frac{1}{K}\right)$ at the surface in contact with the lower plate. The net charge on the lower plate is $-Q+Q\left(1-\frac{1}{K}\right)=-\frac{Q}{K} .$ Consider the equilibrium of the liquid in the volume $A B C D$. The forces on this liquid are
(a) the force due to the electric field at $C D$,
(b) the weight of the liquid,
(c) the force due to atmospheric pressure and
(d) the force due to the pressure of the liquid below $A B$.
As $A B$ is at the same horizontal level as the outside surface, the pressure here is the same as the atmospheric pressure. The forces in (c) and (d), therefore, balance each other. Hence, for equilibrium, the forces in
(a) and (b) should balance each other.
The electric field at $C D$ due to the charge $Q$ is
$$
E_{1}=\frac{Q}{2 A \varepsilon_{0}}
$$
in the downward direction. The field at $C D$ due to the
charge $-Q / K$ is
$$
E_{\mathrm{z}}=\frac{Q}{2 \mathrm{A\varepsilon}_{0} \mathrm{~K}}
$$
also in the downward direction. The net field at $C D$ is
$$
E_{1}+E_{2}=\frac{(K+1) Q}{2 A \varepsilon_{0} K}
$$
The force on the charge $-Q(1-\frac{1}{K})$ at $C D$ is
$$
\begin{aligned}
F=Q\left(1-\frac{1}{K}\right) \frac{(K+1) Q}{2 A \varepsilon_{0} K} \\
=\frac{\left(K^{2}-1\right) Q^{2}}{2 A \varepsilon_{0} K^{2}}
\end{aligned}
$$
in the upward direction. The weight of the liquid considered is $hA\rho g$. Thus,
$$
h A \rho g =\frac{\left(K^{2}-1\right) Q^{2}}{2 A \varepsilon_{0} K^{2}}$$
or,
$$h =\frac{\left(K^{2}-1\right) Q^{2}}{2 A^{2} K^{2} \varepsilon_{0} \rho g}
$$
Solution 2:
Alternative method based on energy.

Suppose the liquid rises by a distance $h .$ Then let us calculate the extra energy of the liquid as a sum of polarization energy and the ordinary gravitational energy. The latter is $$\frac{1}{2} h \cdot \rho g \cdot S h=\frac{1}{2} \rho g S h^{2}$$
If $\sigma$ is the free charge surface density on the palate, the bound charge density is, from the previous problem,
$$
\sigma^{\prime}=\frac{\varepsilon-1}{\varepsilon} \sigma
$$
This is also the volume density of induced dipole moment i.e. Polarization. Then the energy is, as before
$$
-\frac{1}{2} \cdot \sigma^{\prime} E_{0}=\frac{-1}{2} \cdot \sigma^{\prime} \frac{\sigma}{\varepsilon_{0}}=\frac{-(\varepsilon-1) \sigma^{2}}{2 \varepsilon_{0} \varepsilon}
$$
and the total polarization energy is
$$
-S(a+h) \frac{(e-1) \sigma^{2}}{2 \varepsilon_{0} \varepsilon}
$$
Then, the total energy is
$$
U(h)=-S(a+h) \frac{(\varepsilon-1) \sigma^{2}}{2 \varepsilon_{0} \varepsilon}+\frac{1}{2} \rho g S h^{2}
$$
The actual height to which the liquid rises is determined from the formula
$$
\frac{d U}{d h}=U^{\prime}(h)=0 $$This gives:
$$h=\frac{(\varepsilon-1) \sigma^{2}}{2 \varepsilon_{0} \varepsilon \rho g}
$$

This is not a homework question but I just want clarity.
Please explain why the answers are different in a simple language so that even a high school student can understand.
Thanks
 A: Short answer: the first solution makes the incorrect assumption that the pressure in the liquid between the plates is equal to the atmospheric pressure at height $a$.
The second solution is written in an incomplete and confusing fashion but is ultimately sound and gives the correct answer.
I want to give all the credit for setting me on the right path to this comment by @napstablook, which also contains a link to a better version of the second solution:

Have a look at this: I think the half factor comes from there. And it is easily derivable that $P= -σ′$. And I don't think the energy is missing the part you mentioned. That part implies the work done by the electric field of surface charge on itself! the external electric field only does work to polarize this dipole. It seems to me that this new $E$' originated due to the work done on the bound charges by the external electric field. Also, there is an alternative solution given that agrees with the 2nd answer.

Now, about the mistake in the first solution. Let's call the height of the liquid outside the capacitor the "base height", aka $a$. Outside the capacitor, the pressure at the base height is of course equal to the atmospheric pressure. The first solution assumes the same holds inside the capacitor, i.e. that the pressure at the base height is also equal to the atmospheric pressure. But that's an incorrect assumption.
In fact the pressure there is larger. Inside the capacitor, the pressure is equal to the atmospheric pressure at the interface between the liquid and air.
How can the pressure inside the capacitor at the base height be bigger than outside the capacitor at the same height? Wouldn't the liquid start flowing out of the capacitor horizontally, to move from higher to lower pressure? No, for the exact same reason that the water from the bottom of a lake, where the pressure is high, doesn't flow upward to move from higher to lower pressure.
Why doesn't it do that in the lake example? Because the upward force on some small volume element of water due to the pressure differential is compensated by a downward force, namely force of gravity.
In the capacitor situation, the reason the liquid doesn't start flowing out of the capacitor horizontally is analogous. Consider a small volume element of the liquid somewhere near the edge of the capacitor, which experiences the pressure differential pushing it outward. That outward force is compensated by a certain inward force.
What is that inward force? It's the force responsible for the whole phenomenon of the liquid being sucked in and rising inside the capacitor. It's the force with which the non-uniform electric field near the capacitor’s edge (called the fringe/fringing field) acts on the dielectric material of the liquid, polarizing it and pulling it inside the region with the stronger field, i.e. inside the capacitor.
ADDENDUM: another good comment by @napstablook, which expresses several questions that I'm sure many other people will have:

Can you please elaborate why the pressure is larger? @ReasonMeThis the answer looks more focused on what happens if pressure is larger.

This follows immediately from the balance of forces considerations that I described in my answer. Let's look at the lake example again, what did we see there? We saw that for a small volume element of water not to flow (i.e. for the situation to be static), the force due to the pressure differential must be equal and opposite to the external force (gravity in this case).
The same is true for our dielectric liquid. We know that near the edge of the capacitor there is an external force due to the non-uniform electric field (called the fringe/fringing field) trying to pull the liquid inside the capacitor. (How do we know this? See below.) Here again, for the situation to be static the force due to the pressure differential must be equal and opposite to this external force. Therefore the pressure differential must create an outward force, which means the pressure is larger inside the capacitor than outside (at the same height).

Also how can you calculate this extra pressure we need to balance (without resorting to the 2nd method.)

If we already knew the height of the liquid inside the capacitor (which we of course don't until we solve the problem!) then we could use the fact that the pressure must be equal to the atmospheric pressure at the interface between the liquid and air. Without knowing that height it would be extremely difficult to calculate this extra pressure, because it's very hard to deal with the field at the edge of the capacitor.
This is why the approach of the first solution, dealing with forces instead of energies, is actually completely infeasible for this problem. You might find this surprising, but actually if you for example look up the far more standard problem of finding the inward pulling force on a slab of dielectric partially inserted into a capacitor, you'll find that it's always solved with the energy approach. And you'll find discussions about the infeasibility of using the force approach because the fringe fields are so complicated.
And, just to make it explicit, how do we know there even is an inward pulling force to begin with? From energy considerations. Because the dielectric lowers the electric energy density by partially screening the electric field, the overall energy is lower if the dielectric is inserted more fully.

One final thing : 'It's the force with which the non-uniform fringe field near the capacitor’sedge acts on the dielectric material of the liquid, polarizing it and pulling it inside the region with the stronger field." I don't get this last statement.It is because the meaning of fringe field is not cleartome (is there a source you can cite for me to check?)

As I mentioned above, the non-uniform electric field near the edges of a capacitor is called the fringe/fringing field. It is that field that is responsible for pulling dielectrics into the capacitor. I know this seems a bit counterintuitive. But we can see that this must be so, because in order to have an inward pulling force on a dielectric the electric field must have a horizontal component (I am imagining that the capacitor plates are horizontal), but the field inside the capacitor far from the edges is vertical!
