I think I got what it's trying to say.
It basically is trying to say that the electric field at the surface of a metal is always perpendicular to the surface locally.
And it goes on to try to give a justification for it.
Basically the closed loop integral is zero for electrostatic fields in short.the rest comes naturally .
Remember this equation .The closed path integral of work done by an electric field is zero.
In simple works if you take a charge and move it around in space in any manner and at the end come back at the same point where you started then you have done net zero work.
This is KEY in understanding this 'excercise' that your book is doing of rectangle paths etc.
- remember V is basically electric work done in moving a charge of unit magnitude.
-remember that E is basically Force on a charge (a unit charge)
If the whole exercise is confusing then I suggest you that you revisit the definitions of ** E & V**
Think of work when you see V and think of force when you see E.
look at this figure it's from the book
This is the text associated with this 'excercise' from the book (an earlier edition ) with some additions by me in bold and some other cosmetic changes
Remember the Rectangular path is of YOUR choosing!!! YOU choose it in a way for your requirement or convenience. You can choose any path and the formula will hold good
you REALLY REAlly need to have a good understanding of work and forces to get your head around this
At each crossing of an equipotential and a field line，the two are perpendicular .When charges at rest reside on the surface of a conductor，the electric field just outside the conductor must be everywhere perpendicular to the surface。To prove this，we imagine transporting a test charge q' around the loop abcda in Fig。26-9a . The segments be and da（and the associated work）may be made arbitrarily small(because the length is very small Think F.s ie work)，and no work is done in the segment cd because，as already shown，the field is zero everywhere inside the conductor。If the field just outside the conductor has a component E，parallel to the surface，this component does work equal to qlE(//)
that is the parallel component of e vector at the surface (if any) does work on the test charge q' equal to q' times l(length of ab) times parallel component of E at surface E(//)
But then the net work done in the displacement ab is different from that in the displacement adcb，** if you add the contributions from ab adcb paths then the net work will come out to be non zero which will conclude that the ** force field is not conservative。To avoid this contradiction，we must conclude that there cannot be a component of
E parallel to the surface，and that E is therefore perpendicular to the surface。It also follows that，when all charges are at rest，the surface of a conductor is always an equipotential surface，Figure 26-9b illustrates these conclusions。Field lines（color）and equipotentials black）are shown.