# Electric field lines: and nature in presence of a conductor

Long cylindrical shell carries positive surface charge σσ in the upper half and negative surface charge −σ−σ in the lower half. The electric field lines around the cylinder will look like figure given in: (figure are schematic and not drawn to scale)  Note: not a homework question another exam question. (there were 4 options and I am confused between these two )

So my approach was: 1) the field lines go from positive to negative charge 2) they must be tangential to surface of cylindrical shell 3) their shape- this is where I am confused.

in what form must they are? and why is the second figure wrong?

a) due to that sudden change in direction? if yes then how is this related to the nature of electric field lines

Also, can someone cite some study material where I can find other problems related to this type and particularly interaction of field lines with metals

Like: when charge kept inside a body with a non-uniform cavity.

## 1 Answer

The problem is the discontinuity in your field lines, as you suggest. Unless the electric field goes across a charged surface it should be continuous and its derivative should also be continuous. Basically the field lines must be smooth unless there is charge near that field line.

I'm not sure how much calculus you know but this webpage has a nice derivation that continuity of the field components parallel to an interface (as your drawing suggest) must be $0$ in the static case (with no change in magnetic fields). (see also here)

The digest is that the condition you are looking for is encapsulated in $\oint \vec E\cdot d\vec \ell=0$ in the static case. In your specific case one takes the contour to be a frame with sides parallel to the midplane.

• Sorry my friend but I didn't understand this part------- this webpage has a nice derivation that continuity of the field components parallel to an interface (as your drawing suggest) must be in the static case (with no change in magnetic fields) – Aditya Sher Mar 19 '17 at 16:40
• @AdityaSher Your drawings are clearly for electrostatics where there is no moving charges and magnetic fields. Thus the result of linked can be applied by setting $\vec B=0$ in the argument, hence where the right hand side of Eq(635) is $0$. – ZeroTheHero Mar 19 '17 at 16:56