# The consequence of electric field lines not being perpendicular to conductor (assuming a spherical one)

The following statement is from University Physics (chapter "Electric Potential"):

"In particular, at any point just inside the surface the component of E vector tangent to the surface is zero. It follows that the tangential component of E vector is also zero just outside the surface. If it were not, a charge could move around a rectangular path partly inside and partly outside and return to its starting point with a net amount of work having been done on it."

I couldn't understand why would the charges move in rectangular partly in and then partly out and even the part that says net amount of work is done cause conservative forces end up being zero if it returns to it's original point.

## 2 Answers

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**

• You may also substitute V with W(work)

• and E with F (force)if that helps

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.

• it's 14th edition page number 770-771. – suiz Feb 14 '18 at 13:30
• By who? The writer and publisher – Shirish Srivastava Feb 14 '18 at 13:34
• Young and Freedman – suiz Feb 14 '18 at 13:35
• I actually have this but the 6ed. Btw did it clear your doubt? – Shirish Srivastava Feb 14 '18 at 13:36
• I got it. I only thing I was left confused was why the rectangle, but you have answered it. :) – suiz Feb 15 '18 at 15:32

I think that this is the text referred to by the OP. The idea is that if a charge moves around a loop and returns to its starting point no work will be done - the electrostatic force is a conservative force.

So if a charge experiences a force on one side of the loop where there is an electric field and work is done as the charge moves on that side of the loop and then no work is done as the charge moves along the other side of the loop where there no force (electric field) when the charge returned back to where it started from work had been done to move the charge around the loop.
This is at variance to the idea that the electrostatic force is conservative.

• Could you explain why it moves in a loop (a rectangular one specifically), if I understood your loop correctly? Or are you assuming the conductor as a loop or ring? Please clarify. – suiz Feb 14 '18 at 13:31
• But could you explain about how the path turns out to be a rectangle. I couldn't understand it because I found no way that the electric field will lead to a rectangular path for the charge to move (when the electric field has tangential component). – suiz Feb 14 '18 at 13:42
• @suiz It is just a convenient path to analyse as evaluating the work done alone a straight line which is parallel to an electric field is easy to evaluate. – Farcher Feb 14 '18 at 13:42
• @suiz The rectangular path is an imaginary path along which you “carry” the charge. – Farcher Feb 14 '18 at 13:44
• Ok, so if the electric field has a tangential component work is done to move in along one side of loop but since the force is conservative net work on the charge should be zero when it reaches the starting point but since there is no electric field inside the conductor, there won't be any force which means that it will defy law of conservation. Hence, electric field can't have tangential component. It's almost same what you have written. But got this point. You cleared it. – suiz Feb 15 '18 at 6:50