Case: Read the passage and answer the question.
Conductors contain mobile charge carriers. In metallic conductors, these charge carriers are electrons. In a metal, the outer (valence) electrons part away from their atoms and are free to move. These electrons are free within the metal but not free to leave the metal. The free electrons form a kind of ‘gas’; they collide with each other and with the ions and move randomly in different directions. In an external electric field, they drift against the direction of the field. The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions. A dielectric (an insulator) behaves differently. Its charges are held to their atoms or molecules and cannot move freely through it. An external electric field can only shift these charges a little or turn the molecules. The dielectric is then said to be polarised.
Which of the following is the correct expression for the electric field at the surface of a charged conductor?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \vec{E}=\displaystyle\frac{\sigma}{\varepsilon_0}\,\hat{n} \)
Option (1) \( E=q/r \) — Incorrect. This expression has no factor \( 1/4\pi\varepsilon_0 \) and no inverse-square dependence on distance. The field at a conductor’s surface depends on the local surface charge density. It does not depend on a total charge and a distance.
Option (2) \( \vec{E}=\displaystyle\frac{\sigma}{\varepsilon_0}\,\hat{n} \) — Correct. Just inside a conductor the electrostatic field is zero, and just outside it is normal to the surface. Gauss’s law, applied to a small pill box across the surface, gives the magnitude \( \sigma/\varepsilon_0 \). Here \( \sigma \) is the surface charge density, and \( \hat{n} \) is the outward unit normal.
Option (3) \( E=\varepsilon_0/\sigma \) — Incorrect. This is the correct ratio turned upside down. The field grows with the surface charge density, \( E\propto\sigma \). Also, \( \varepsilon_0/\sigma \) does not have the unit of electric field.
NCERT (2026-27):
Electrostatics of Conductors – “5. Electric field at the surface of a charged conductor
\( \displaystyle \mathbf{E}=\frac{\sigma}{\varepsilon_0}\,\hat{\mathbf{n}} \)\( \qquad (2.35) \)
where \( \sigma \) is the surface charge density and \( \hat{\mathbf{n}} \) is a unit vector normal to the surface in the outward direction.”