Electrostatics - Long Question Answer | Physics Grade 11 Solution

Long Question Solution

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Electrostatics - Long Question Answer | Physics Grade 11 Solution

Electrostatics - Long Question Answer 4

Question: Use Gauss law to find the electric field intensity due to a plane charged conductor.
Ans: Consider a plane conductor with the uniform surface charge density, σ. To find the electric field intensity at a point P outside the conductor. Let’s draw a cylinder of cross-section A such that the top of it lies at point P and the bottom part of it lies in the plane of the conductor. Then by symmetry, the electric field intensity at every point on the cross-sectional area A should be same. So, the flux passing normally through the surface area A is given by φ = E * A
Then net charge Q enclosed by the Gaussian surface is q = σ * A
Then for Gauss law, the total flux passing through the cylinder is
Φ = q/ϵ0
or, E * A = σA/ϵ0
or, E = σ/ ϵ0
.: This is required electric field intensity due to a plane charged conductor.

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