jangyasinniTeacher

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To find the electric field ((E)) resulting from a given charge density ((sigma)) at the boundary of two media, such as at the boundary with air, you can use Gauss’s law. In this context, assuming the surface charge density ((sigma)) is uniform and considering the medium is air (which can be treated as a vacuum for most practical purposes in electromagnetism), we consider the formula derived from Gauss’s law for a planar symmetry.

For a surface charge density, the electric field just outside a charged surface can be given by:

[E = frac{sigma}{2epsilon_0}]

where

– (sigma = 10^{-9} , text{C/m}^2) (surface charge density),

– (epsilon_0 = 8.85 times 10^{-12} , text{C}^2/text{N}cdottext{m}^2) (permittivity of free space).

Substituting the given values:

[E = frac{10^{-9}}{2 cdot 8.85 times 10^{-12}}]

[E = frac{10^{-9}}{1.77 times 10^{-11}}]

[E = frac{1}{1.77} times 10^{2}]

[E approx 56.5 , text{N

c

Explanation: It is the conductor-free space boundary. At the boundary, E = ρ/εo. Put ρ =

10

-9 and εo = 10

-9

/36π. We get E = 36π units.