My book says that the electric field due to infinite long plates doesn't depend on 'r', so does it mean that no work is done in moving a point charge towards or away from it?
Also considering the situation of parallel plate capacitors , even here the electric field between the 2 plates doesn't depend on the distance from plates but still there is a potential difference between the plates that does depend on the distance between the 2 plates. How is this possible ??
so does it mean that no work is done in moving a point charge towards or away from it?
No. A charge under the influence of an electric field has a force exerted on it.
So to move the charge work needs to be done.
The electric field strength is equal to minus the potential gradient.
So if there is an electric field the potential must change.
Now when the separation of the plates of a parallel plate capacitor is changed there are two possibilities.
The capacitor is not connected to anything so the charges on the plates stay the same which means the the electric field $E$ between the plates must stay the same.
However to separate the plates external work must be done and the potential difference between the plates will increase as $E = (-) \frac V d$ where $V$ is the potential difference between the plates and $d$ is the separation of the plates.
If a battery is placed across the capacitor it will maintain a constant potential difference $V$ across the plates.
So if the separation of the plates is increased the electric field $E = \frac V d$ must decrease.
The electric field decrease because the electrons flow from the negative plate of the capacitor into the negative terminal of the battery and out of the positive terminal of the battery onto the positive plate of the capacitor.
Thus the amount of charge stored on the capacitor has decreased.
External work must be done to separate the plates.
Your book just means that the electric field between a pair of parallel plates is uniform, by that it is both constant in direction and value (this is actually only approximately true in practice, there is a small variation).
There is a potential difference between the plates because there is an electric field between the plates. Because the field is uniform it means that the relationship between voltage difference, $V$, separation of plates $d$ and size of electric field is especially simple, namely $E = \frac{V}{d}$.
