The Field in a Parallel Plate Capacitor (E = V / d and the Number of Field Lines)

The field between the plates has the same strength everywhere

The field between two facing plates runs straight from the positive plate to the negative one. Anywhere in the middle it has the same direction and the same strength, and a field like that is called uniform.

Being uniform is what the equal arrow lengths in the figure show. The field around a point charge weakens with distance; between plates it does not.

Only at the edges does it fail, where the field lines bulge outward. If the gap is small enough compared with the width of the plates, that edge effect can be ignored.

The field is E = V / d

Crossing a uniform field from the positive plate to the negative one over a distance dd, the potential drops by VV. Since the field is uniform the drop is steady, so the field strength is E=VdE = \dfrac{V}{d}.

Hold the voltage and narrow the gap, and the same drop is made over a shorter distance, so the field is stronger. The narrow gap is on the left of the figure, the wide one on the right.

The capacitance follows from the same picture. With plate area SS, C=ε0SdC = \varepsilon_0\dfrac{S}{d}: the narrower the gap and the wider the plates, the larger the capacitance.

The number of field lines is proportional to the charge

Field lines are a way of making a field visible, and the number of them carries no meaning on its own. But once you decide to draw the number in proportion to the charge, quantities can be read off the picture.

That convention comes from the Gauss law, which says that the total number of lines through a closed surface is fixed by the charge enclosed by it.

In the figure, changing the charge changes the number of lines. Changing the gap does not. What the gap changes is the voltage, not the charge.

Disconnect first, and changing the gap moves the voltage

Charge the capacitor, then take it off the battery, and the charge has nowhere to go. Widen the gap now and QQ does not change. Neither does the number of field lines.

What changes is the voltage. The field is E=Qε0SE = \dfrac{Q}{\varepsilon_0 S}, fixed by the charge alone, so it stays as it was, and V=EdV = Ed rises with the gap.

Leave the battery connected and the answer reverses. There VV is held fixed, so widening the gap lowers CC and, by Q=CVQ = CV, the charge falls. Which quantity is held fixed decides the answer, so settle that before anything else.