# A dielectric table is being inserted between a plate capacitor and $\triangle U<0$ how to deduce the table is attracted to the plates?

I am practicing for an exam in my Physics $2$ course.

One of a previews exam questions described a plate capacitor and asked to calculate the initial energy $U_{0}$, then a dielectric table was inserted between the plates and I was asked to calculate the energy, $U_{1}$, at this state.

I got that there was less energy in the final state than there was at the initial state, i.e $$\triangle U=U_{1}-U_{0}<0$$

The question asked if the table is attracted to the plates or repealed from it, and the answer claimed that since there was less energy at the final state then it means that the table is attracted to the plates.

I lack intuition on this and I don't understand how the conclusion was made.

I tried to think it with terms of work done by the electrical field: If I assume that the table is attracted to the plates then I think that the work is negative (since I don't have to do actual work because there is an attraction),

and since there was negative work done $W<0$ I have $$U_{1}=U_{0}+W<U_{0}$$ The other case leads to $U_{1}>U_{0}$ and so I get that the first is the one that occurs.

But I am not to sure about this argument since $$W=\int F\cdot dl$$ and since there is an attraction the direction of the path is the same as the direction of the force and so $W>0$.

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The change in the potential energy of the electric field $\Delta U$ is equal to the negative of the work done $W$ by the electric force.
You have $$\Delta U = -W$$ $$U_1=U_0-W<U_0$$ $$\therefore W>0$$
And as $$W=\int F\cdot dl>0$$, we can say that $F$ acts in the same direction as $dl$, i.e it is attractive.
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