# Why are field lines always perpendicular to conducting surfaces?

## Answer #1

Charges in **conductive surfaces** (we call them surface charges) can move *freely* along the surface. This is what characterizes conductors. This charges cannot move perpendicular to the surface because otherwise they would escape the conductor.

Let us assume that the field lines do NOT point perpendicularly out of the surface of the conductor. Accordingly, the associated **electric field vectors** \(\boldsymbol{E}\) (and hence the electric forces \( \boldsymbol{F} = q \, \boldsymbol{E}\) on the charges) will not point perpendicularly out of the surface. Thus, the vector \(\boldsymbol{E}\) at the surface would point in any direction, as illustrated in Illustration 1.

The field vector \(\boldsymbol{E}\) can be mathematically split into a perpendicular \(\boldsymbol{E}_{\perp} \) and parallel \(\boldsymbol{E}_{||} \) component. The parallel \(\boldsymbol{E}_{||} \) field component will cause the surface charges to move *along the surface* because \(\boldsymbol{E}_{||} \) is pointing along the conducting surface. The free charges will continue to move along the surface until they experience no force in that direction. The force parallel to the surface is then zero. Consequently, the paralel field component also disappears: \(\boldsymbol{E}_{||}=0 \). Only a force on the charges perpendicular to the surface remains and thus also the perpendicular field component \(\boldsymbol{E}_{\perp} \). This field component cannot be compensated, because the charges would have to move out of the surface.