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How does a capacitor behave when all of its plates are negatively charged?

The plates of the capacitor are negatively charged. So the potential difference between them is also negative. Then if a neutral object is placed between them, the electric field lines on the walls are such that they are opposite the electric field lines on the object.
The answer to this question was given to be zero. But I don’t think this is the case. When the distance between the negative plates is very small or when the object is placed right in the middle of the two plates, I think the potential difference between the plates due to the object is negative (which means it is like a negative charged object with all the electric field lines pointing towards it) which means the field lines in the gap is the same as the field lines on the object. So how does this not make the potential difference between the plates due to the object, negative?

A:

But I don’t think this is the case. When the distance between the negative plates is very small or when the object is placed right in the middle of the two plates, I think the potential difference between the plates due to the object is negative (which means it is like a negative charged object with all the electric field lines pointing towards it) which means the field lines in the gap is the same as the field lines on the object. So how does this not make the potential difference between the plates due to the object, negative?

Why would that make a difference? The potential difference between the two plates due to the object is zero because the net electric flux through the plates is zero. However, this net electric flux is not zero for the sum of the electric fluxes between the object and the two plates.
Here is a graphical depiction of the situation:

The net electric flux through the plates due to the object is zero, so $\int \vec E\cdot d\vec S=0$. The electric flux between the object and plate 1 is the negative of the electric flux between the object and plate 2, because the positive charge of the object is moved towards plate 1, and the positive charge of the other plate is moved towards the object. These two fluxes combine to give \$\int \vec E\cdot d\vec S=0

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