A planar insulating sheet with an infinitesimal thickness has charge per unit area σ and is parallel to the x−z plane, as shown. (The arrows indicate that the object extends for a long distance in the horizontal plane.) By the symmetry of the problem, the electric field may be expressed as E⃗ =Ej^. This problem relates the change in the electric field from the lower side to the upper side of the insulating sheet to the surface charge density.

a) Enter a vector expression, in Cartesian unit-vector notation, for the electric field at points just above the planar insulating sheet.
b)Enter a vector expression, in Cartesian unit-vector notation, for the electric field at points just below the planar insulating sheet.

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oladayoademosu oladayoademosu · Answer 1 · 2022-04-03T01:06:23+02:00

a. The electric field above the planar insulating sheet is Ej = (σ/ε₀)j

b. The electric field below the planar insulating sheet is -Ej = -(σ/ε₀)j

The electric field above the planar insulating sheet is Ej = (σ/ε₀)j

Using Gauss' law ε₀∫E.dA = q where

ε₀ = permittivity of free space,
E = electric field vector normal to the surface above the planar insulating sheet = Ej,
dA = normal vector to surface above the planar insulating sheet = dAj,
q = charge enclosed by Gaussian surface on planar insulating sheet = σA where
σ = surface charge density and
A = area of surface

Now, ε₀∫E.dA = q

ε₀∫Ej.dAj = σA

ε₀∫EdA = σA

ε₀E∫dA = σA

ε₀EA = σA

ε₀E = σ

E = σ/ε₀

So, the electric field above the planar insulating sheet is Ej = (σ/ε₀)j

The electric field below the planar insulating sheet is -Ej = -(σ/ε₀)j

Using Gauss' law ε₀∫E.dA = q where

ε₀ = permittivity of free space,
E = electric field vector normal to the surface below the planar insulating sheet = -Ej,
dA = normal vector to surface below the planar insulating sheet = -dAj,
q = charge enclosed by Gaussian surface on planar insulating sheet = σA where
σ = surface charge density and
A = area of surface

Now, ε₀∫E.dA = q

ε₀∫-Ej.-dAj = σA

ε₀∫EdA = σA

ε₀E∫dA = σA

ε₀EA = σA

ε₀E = σ

E = σ/ε₀

So, the electric field below the planar insulating sheet is -Ej = -(σ/ε₀)j

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