Electrostatics> why is the electric field perpendicular t...

why is the electric field perpendicular to the surface of a conductor?

sushmitha , 5 Years ago

Grade 12

2 Answers

Arun

Last Activity: 5 Years ago

Surface of the conductor is an equipotential surface for static distribution of charge on it. Had it not been so, then charge would flow between points with potential difference. But, then we can not have static charge on the surface.

Now, suppose electric field lines are not perpendicular to the local surface of the conductor . Then electric field will have some component parallel to the surface. This component will give potential difference dV=-E( parallel).dl between the nearby points. Then, the surface is not equipotential surface. So, for surface to be equipotential surface for static charge distribution the field lines at the surface are perpendicular to the surface.

Khimraj

Last Activity: 5 Years ago

First you need some vectors.

If a vector, say a force vector, is making an angle of theta with the surface of an object, then that force will have a perpendicular component Fsin thetha and a tangential or parallel component (parallel to the surface) F cos theta.

These forces will try to accelerate the particles along and perpendicular and tangential directions with respect to the ground.

Now coming back to your question. If a field like is making an angle theta with the surface of a charged conductor, then that field line will try to accelerate the local charge (the charges present on the surface near that field line) tangentially as well as perpendicularly.
Now the charge cannot get accelerated perpendicularly as other forces will try to pull it back and cancel the forces out.

However the tangential component would accelerate the charges and that would make them move through the conductor. But in electrostatics, charges are supposed to be at rest.

Thus we conclude that eventually when electrostatic conditions have reached, the field lines must be perpendicular to the surface everywhere.

And yes before you ask, this is only valid as long as we are in electrostatic conditions, as you can easily make out from my argument.