Q.Applying Gauss's theorem, find the electric field at any point due to a uniformly charged infinite conducting thin plate.

Using a Gaussian pillbox straddling the conductor's surface, with the field inside the conductor being zero, Gauss's law gives E = σ/ε₀ just outside the plate.

Consider an infinite thin conducting plate carrying a uniform surface charge density $\sigma$. Since it is a conductor, the electric field inside the conductor is zero, and all the field emerges perpendicular to the two outer surfaces.

To find the field just outside, construct a small cylindrical Gaussian pillbox ("pillbox") that straddles the surface: one flat face of area $A$ lies just outside the conductor (in the field region), and the other flat face lies just inside the conductor (where $E = 0$). The curved side is perpendicular to the field lines, so it contributes no flux.

Applying Gauss's law, $\oint \vec E\cdot d\vec A = \dfrac{q_{enc}}{\varepsilon_0}$:

• Flux through the outside face: $E\cdot A$ (field is normal to the surface).
• Flux through the inside face: $0$ (field is zero inside the conductor).
• Flux through the curved side: $0$.
• Charge enclosed: $q_{enc} = \sigma A$ (the surface charge under the pillbox).

So: …