Why must electrostatic field be normal to the surface at every point of a charged conductor?

Open in App

Solution

The charges on the surface of a conductor are static, that is, they do not experience any force. It is only when there is no component of electric field along the surface of charged body.
Hence electric field is normal to the surface.

Suggest Corrections

Similar questions

Q. Why must electrostatic field at the surface of a charged conductor be perpendicular to every point on it?

Q. Why must the electrostatic potential inside a hollow charged conductor be the same at every point?

Q. Assertion :Net charge given to a conductor resides on its outer surface under electrostatic conditions. Reason: Electric field lines do not enter a conductor under electrostatic conditions.

Q. Why should electrostatic field be zero inside a conductor?

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

Where is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

Join BYJU'S Learning Program

Why must electrostatic field be normal to the surface at every point of a charged conductor?

The charges on the surface of a conductor are static, that is, they do not experience any force. It is only when there is no component of electric field along the surface of charged body.Hence electric field is normal to the surface.

The charges on the surface of a conductor are static, that is, they do not experience any force. It is only when there is no component of electric field along the surface of charged body.
Hence electric field is normal to the surface.